course Mth 152 Here is the rest of the assignments. I went to take the tests and the lab was closed. I will try to take them tomorrow and if I can't, i will have them done on Monday. ִՉoDMj`assignment #009
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23:59:08 Query 12.4.3 P(2 H on 3 flips)
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RESPONSE --> 3/8=0.375
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23:59:19 ** On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of there 8 possibilities, only 3 of them have two Heads. Thus the probability is 3 / 8. You can get this result without listing. There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. If you have n flips, there are C(n,r) ways to get r Heads. This number appears in the n+1 row, as the r+1 entry, of Pascal's triangle. **
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RESPONSE --> ok
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00:01:52 What is the significance of .5^2 * .5 for this question?
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RESPONSE --> it is the probability of getting heads twice in a row
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00:01:59 ** .5^2 is the probability of getting Heads twice in a row. .5 is the probability of a Tails. .5^2 * .5 is therefore the probability of getting HHT. Since the probabilities are independent, you have the same probability of getting two Heads and a Tail in some different order. Since there are C(3,2) possible orders for 2 Heads on 3 coins, the probability of getting 2 Heads and one Tail is C(3,2) * .5^2 * .5 = 3 * .125 = .375, the same as the 3/8 we obtained by listing. **
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RESPONSE --> ok
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00:03:43 Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> 7/8= 0.875 this is the probability of getting no heads on three flips
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00:03:51 ** Probability of getting no heads on three flips is P(TTT) = .5 * .5 * .5 = .125, or 1/8, obtained by multiplying the probability of getting a tails for each of 3 independent flips. Subtracting this from 1 gives .875, or 7/8. **
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RESPONSE --> ok
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00:06:22 Query 12.4.15 P(3 H on 7 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> c(7,3)=7*6*5/3= 35 ways 35/128
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00:06:28 ** There are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to choose three of the 7 `positions' for Heads on 7 flips. So there are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to get three heads on 7 flips. The probability of any of these ways is (1/2)^3 * (1/2)^4 = 1 / 2^7 = 1 / 128. The probability of 3 Heads on 7 flips is therefore 35 * 1/128 = 35 / 128. **
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RESPONSE --> ok
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00:08:39 Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die
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RESPONSE --> c(3,1)*1/6*(5,6)^2=4*1/6*25/36=100/216=25/72
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00:08:43 ** To get 1 success on 3 tries you have to get 1 success and 2 failures. On any flip the probability of success is 1/6 and the probability of failure is 5/6. For any ordered sequence with 1 success and 2 failures the probability is 1/6 * (5/6)^2. Since there are C(3,1) = 3 possible orders in which exactly 1 success can be obtained, the probability is C(3,1) * 1/6 * (5/6)^2 = 4 * 1/6 * 25 / 36 = 100 / 216 = 25 / 72. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/6, prob of failure q = 1 - 1/6 = 5/6, n = 3 trials and r = 1 success. **
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RESPONSE --> ok
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00:10:52 Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability?
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RESPONSE --> 1/3 (1/3)^7*(2/3)^3 c(10,7) 320/19683=0.016
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00:10:56 ** The probability of a correct answer from a random choice on any single question is 1/3. For any sequence of 7 correct answers and 3 incorrect the probability is (1/3)^7 * (2/3)^3. There are C(10,7) possible positions for 7 correct answers among 10 questions. So the probability is C(10,7) * (1/3)^7 * (2/3)^3 = 320/19683 = 0.0163 approx. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/3, prob of failure q = 1 - 1/3 = 2/3, n = 10 trials and r = 7 success. ANOTHER SOLUTION: There are C(10,7) ways to distribute the 7 correct answers among the 10 questions. The probability of any single outcome with 7 successes and 3 failures is the product of (1/3)^7, representing 7 successes, and (2/3)^3, representing 3 failures. The probability of exactly seven correct questions is therefore prob = C(10,7) * (2/3)^3 * (1/3)^7 . **
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RESPONSE --> ok
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00:12:11 Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each
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RESPONSE --> 0.448
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00:12:18 ** The probability of 0 side effects is C(8,0) * .7^8. The probability of 1 side effect is C(8,1) * .7^7 * .3^1. The probability of 2 side effects is C(8,2) * .7^6 * .3^2. The sum of these two probabilities is the probability that two or fewer patients will have side effects. We subtract this probability from 1 to get the probability that more than 2 will experience side effects. The result is approximately .448. DER**
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RESPONSE --> ok
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00:13:10 Query 12.4.48 P(4 th child is 1 st daughter) What is the probability that the fourth child is the first daughter and how did you obtain your result?
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RESPONSE --> 1/2*1/2*1/2*1/2=0.0625
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00:13:16 ** The fourth child will be the first daughter if the sequence is SSSD, S standing for son and D for daughter. The probability of S on any birth is .5, and the probability of G is .5. The probability of SSSD is .5^3 * .5 = .0625 or 1/16. **
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RESPONSE --> ok
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00:15:50 Query 12.4.54 10-step rnd walk, 1 dim; P(6 South) What is the probability of ending up 6 blocks South of the starting point and how did you obtain it?
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RESPONSE --> 1/10*1/6=0.016
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00:15:57 ** To end up 6 blocks South requires 8 steps South and 2 steps North. Thus exactly 8 of the 10 steps must be South, and there are C(10,8) ways for this to happen. The probability of any given combination of 8 South and 2 North is (1/2)^8 * (1/2)^2 = 1 / 2^10 = 1 / 1024. خ܀Jé{җ˨[X͂ assignment #010 cʓ齯̀q} li Liberal Arts Mathematics II 11-28-2006
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00:24:16 Query 12.5.6 fair dice game pays $3 for 6, $2 for 5, $1 for 4. What is a fair price to pay for playing this game?
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RESPONSE --> 3.00 per game
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00:24:30 ** A 1 in 6 chance of getting $3 is worth 1/6 * $3 = $.50 . A 1 in 6 chance of getting $2 is worth 1/6 * $2 = $.33 1/3 . A 1 in 6 chance of getting $1 is worth 1/6 * $1 = $.16 2/3 . The total expectation is $1.00 * 1/6 + $2.00 * 1/6 + $3.00 * 1/6 = $1.00 So a fair price to pay is $1.00 **
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RESPONSE --> ok
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00:28:00 Query 12.5.10 expectation Roulette $1 bet 18 red, 18 black one zero What is the expected net value of a bet on red?
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RESPONSE --> ($1)18/37=0.0486
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00:28:13 ** If your net gain is $1 for a win and -$1 for a loss the expected value is 18/37 * (+1) + 19/37 * (-1) = -$.027. **
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RESPONSE --> ok
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00:30:07 Query 12.5.20 exp sum of 2 of 5 cards 1-5. What is the expected sum of the numbers on the two cards drawn?
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RESPONSE --> 1/5+1/5=2/5
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00:30:17 ** You can't get a sum of 1 on two cards. There is also no way to get a sum of two, since the lowest total possible is 1 + 2 = 3. There are 2 ways to get total 3. You can get 1 on the first and 2 on the second, or vice versa. There are 2 ways to get total 4. You can get 1 on the first and 3 on the second, or vice versa. There are 4 ways to get total 5. You can get 1 on the first and 4 on the second, or vice versa, or 2 on the first and 3 on the second, or vice versa. There are 4 ways to get total 6. You can get 1 on the first and 5 on the second, or vice versa, or 2 on the first and 4 on the second, or vice versa. There are 4 ways to get total 7. You can get 2 on the first and 5 on the second, or vice versa, or 4 on the first and 3 on the second, or vice versa. There are 2 ways to get total 8. You can get 3 on the first and 5 on the second, or vice versa. There are 2 ways to get total 9. You can get 4 on the first and 5 on the second, or vice versa. You can't get more than 9. There are 2+2+4+4+4+2+2 = 20 possibilities, so the probabilities are 2/20, 4/20, 5/20, etc.. The expected sum is therefore 2/20 * 3 + 2/20 * 4 + 4/20 * 5 + 4/20 * 6 + 4/20 * 7 + 2/20 * 8 + 2/20 * 9. This gives 120 / 20 = 6. **
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RESPONSE --> ok
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00:30:23 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ok
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Ƶw׆wǪ assignment #011 cʓ齯̀q} li Liberal Arts Mathematics II 11-28-2006
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00:38:55 **** Query 12.6.6 rnd # table to simulate 50 one-and-one foul shooting opportunities if 70% prob of success; 2 shots Give the results of your tally. How does your empirical probability compare with the theoretical probability?
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RESPONSE --> Pts. Tally 0 2 1 0 2 0
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00:39:10 ** In 1-and-1 shooting you only get a second shot if you make the first. So there are 3 possibilities: miss the first, don't get another shot make the first, get another shot and make it make the first, get another shot and miss it prob of 0 = prob of miss on first shot = .3 prob of 1 = prob of hit on first and miss on 2d = .3 * .7 = .21 prob of 2 = prob ot hit * prob of hit = .49. 'Hits' happen with 70% or .7 probability, misses with probability 30% or 3. The theoretical probability of 2 misses is probability of miss * probability of miss = .30 * .30 = .09. The theoretical probability of 2 miss and 1 hit is probability of miss * probability of hit + probability of miss *hit probability of miss = .30 * .70 + .70 * .30 = .21 + .21 = .42. The theoretical probability of 2 hits is probability of hit * probability of hit = .70 * .70 = .49. Note that these probabilities add up to .09 + .42 + .49 = 1, as they must since these three events cover all possibilities. To use the table, randomly pick a starting point. Let numbers 1-7 correspond to making the free throw, with 8, 9 and 0 corresponding to misses. Go down the list, or across the list in an order you decided before looking at the list. Read two digits from the list and see if they correspond to two 'hits', two 'misses' or a 'hit' and a 'miss'. Record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Read two more digits and record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Continue until you have the required number of results. Tally how many times you got 0 'hits', 1 'hit', 2 'hits' etc.. Any outcome that starts with a 'miss' corresponds to zero point. 'Hit-miss' corresopnds to 1 point and 'hit-hit' corresponds to 2 points. Determine the percent of time you got each number of points, and compare to the theoretical probabilities .09, .42 and .49. *&*& **
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RESPONSE --> ok
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00:43:40 Query 12.6.12 rnd walk start N then right, left or straight with prob 1/2, 1/6, 1/3; 1 st 2 columns of table
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RESPONSE --> 1/2*1/6*1/3=0.02
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00:45:12 ** Your probabilities are given as 1/2, 1/6 and 1/3. These can all be expressed in terms of the common denominator 6: 1/2 = 3/6, 1/6 = 1/6, 1/3 = 2/6. So a move to the right has 3 chances out of 6, a move to the left has 1 chance out of 6 and a move straight has 2 chances out of 6. You can simulate this by letting the three digits 1, 2, 3 stand for a move to the right, the single digit 4 for a move to the left and the two digits 5, 6 for a straight move. The remaining digits 0, 7, 8, 9 don't stand for anything, and if you land on one of these numbers you just move to the next number. So according to your the first two columns of you table, how many times do you move to the right, how many to the left, how many straight and where do you end up? **
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RESPONSE --> 4 lefts, 4 straights back at the beginning?
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00:45:24 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I'm completely confused
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00:45:34 014. `query 14
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RESPONSE --> ok
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00:52:11 query problem 13.1.6 freq dist 35 IQ scores class width 5
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RESPONSE --> I'm not sure if I understand this but if I did it correctly on paper, the graph shows an inrease in IQ scores, starting at 91 and going all the way to 134
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00:52:53 ** These numbers might or might not be completely accurate, but the calculation of the relative frequencies as percents follows from the tallies: 91-95 1 1/50 = 2% 96-100 3 3/50 = 6% 101-105 5 5/50 = 10% 106-110 7 7/50 = 14% 111-115 12 12/50 = 24% 116-120 9 9/50 = 18% 121-125 8 8/50 = 16% 126-130 3 3/50 = 6% 131-135 2 2/50 = 4% **
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RESPONSE --> oh, I see how it goes now
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01:04:26 query problem 13.1.10 stem and leaf for yards gained by 44 rushers
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RESPONSE --> 0 3,7 1 2,6,8,9 2 2,4,5,8,9 3 0,2,3,3,6,6,7,9 4 0,1,2,3,3,5,6,9 5 1,4,5,8 6 0,2,7 7 3,3,9 8 6,8 9 4 10 2 11 2 12 3
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01:04:40 ** In order we get the following: 0 3 7 1 2 6 8 9 2 2 4 5 8 9 9 3 0 2 3 3 6 6 7 9 4 0 1 2 3 3 5 6 9 5 1 4 5 8 6 0 2 7 7 3 3 9 8 6 8 9 4 10 2 11 2 12 3 **
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RESPONSE --> ok
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01:06:26 query problem 13.1.35 empirical probability distribution for letters of alphabet
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RESPONSE --> letter probability A 0.208 E 0.338 I 0.169 O 0.208 U 0.078
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01:06:55 ** The probabilities for A, E, I, O, U are: .08. .13, .065, .08, .03. The sum of these probabilities is .385 The probabilities of the letters given a vowel: .08 / .385 = .208 .13 / .385 = .338 .065 / .385 = .169 .08 / .385 = .208 .03 / .385 = .078 These probabilities total 1. **
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RESPONSE --> OK
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01:07:05 015. `query 15
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RESPONSE --> OK
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01:11:29 query problem 13.2.10 .3, .4, .3, .8, .7, .9, .2, .1, .5, .9, .6
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RESPONSE --> MEAN- 0.52 MEDIAN- 0.5 MODE- 0.3, 0.9
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01:11:54 ** The numbers, in order, are .1, .2, .3, .3, .4, .5, .6, .7, .8, .9, .9 The mean, obtained by adding the 11 numbers then dividing by 11, is .518. The median occurs at position (n + 1 ) / 2 = 6 in the ordered list. This number is .5. Note that there are five numbers before .5 and five numbers after .5. The maximum number of times a number repeats in this distribution is 2. So there are two modes (and we say that the distribution is bimodal). The modes are .3 and .9. **
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RESPONSE --> I miscounted
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01:16:50 **** query problem 13.2.24 more effect from extreme value
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RESPONSE --> the mean would show more of an effect
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01:17:01 ** The mean is drastically affected by the error; correcting the error changes the mean by about 3 units. The median number, however, simply shifts 1 position, changing from 2.28 to 2.39. **
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RESPONSE --> ok
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01:21:43 **** query problem 13.2.30 Salaries 6 @$19k, 8 @ 23k, 2 @ 34.5k, 7 @ 56.9k, 1 @ 145.5k.
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RESPONSE --> the mean is 135,000
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01:22:03 ** IF THERE ARE 28 EMPLOYEES: The totals paid for each salary level are: 6 * $19,500 = $117,000 8 * $23,000 = $184,000 4 * $28,300 = $113,200 2 * $34,500 = $69,000 7 * $36,900 = $258,300 1 * $145,500 = $145,500 The grand total paid in salaries to the 28 employees is therefore $887,000, giving an average of $887,000 / 28 = $31,700. The median occurs at position (n + 1) / 2 = (28 + 1) / 2 = 14.5. Since the 14 th salaray on a list ordered from least to greatest is $23,000 and the 15 th is $28300 the median is ($23000 +$28300) / 2 = $25,650. The mode is 23,000, since this salary occurs more frequently than any other. IF THERE ARE 24 EMPLOYEES: The totals paid for each salary level are: $19,000 * 6 = $114,000 $23,000 * 8 = $184,000 $34,500 * 2 = $69,000 $56,900 * 7 = $398,300 $145,500 * 1 = $145,500 Adding these gives a grand total, which is divided by the number 24 of employees to obtain the mean $37,950. The median occurs at position (n + 1) / 2 = (24 + 1) / 2 = 12.5. Since the $23000 salary covers positions 7 thru 14 in an ordered lise of salaries the median is $23,000. The mode is 23,000, since this salary occurs more frequently than any other.
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RESPONSE --> ok
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01:22:56 **** query problem 13.2.51 mean, med, mode of 0, 1, 3, 14, 14, 15, 16, 16, 17, 17, 18, 18, 18, 19, 20
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RESPONSE --> mean- 13.7 median- 16 mode- 18
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01:23:02 ** The mean is 13.73, obtained by adding together all the numbers and dividing by n = 15. The median is in position (n+1) / 2 = (15+1)/2 = 8 on the ordered list; the 8 th number is 16. The mode is 18, which is the only number occurring as many as 3 times. **
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RESPONSE --> ok
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01:23:07 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ok
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01:23:24 016. `query 16
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RESPONSE --> ok
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01:31:18 query probl 13.3.6 range, std dev of {67, 83, 55, 68, 77, 63, 84, 72, 65}
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RESPONSE --> range-84-55=29 mean- 70.4 standard deviation- 55-70.4= -15.4 63-70.4= -7.4 65-70.4= -5.4 67-70.4= -3.4 68-70.4= -2.4 72-70.4= 1.6 77-70.4= 6.6 83-70.4= 12.6 84-70.4= 13.6
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01:31:42 ** x dev. from mean squared dev. 55 15.4 237.16 63 7.4 54.76 65 5.4 29.16 67 3.4 11.56 68 2.4 5.76 72 1.6 2.56 77 6.6 43.56 83 12.6 158.76 84 13.6 184.96 634 728.08 mean = 634 / 9 = 70.4 std. dev. = `sqrt (728.08 / 8) = 9.54 range = 84 - 55 = 29 **
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RESPONSE --> ok
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01:49:19 **** query probl 13.3.12 freq dist 14,8; 16,12; 18,15; 20,14; 22,10; 24,6; 26,3
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RESPONSE --> 6 4 3 6 12 18 23 mean-10.3 6-10.3=-4.3 18.49 4-10.3=-6.3 39.69 3-10.3=-7.3 53.29 6-10.3=-4.3 18.49 12-10.3=1.7 2.89 18-10.3=7.7 59.29 23-10.3=12.7 161.29
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01:49:33 ** Value freq Value * Freq Dev^2 * F 14 8 112 204.80 16 12 192 112.32 18 15 270 16.80 20 14 280 12.32 22 10 220 86.40 24 6 144 146.40 26 3 78 144.48 Total 68 1296 723.52 Total squared dev is 723.5 so ave squared dev is 723.5 / 68 = 10.6, approx. Std dev is sqrt(ave squared dev) = sqrt(10.6) = 3.3 approx. **
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RESPONSE --> ok
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01:54:14 **** query probl 13.3.18 chebyshev for z=5 What is the least possible number of elements of a sample which lie within 5 standard deviations of the mean?
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RESPONSE --> 1- 2/5= 0.60
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01:54:28 ** The formula 1 - 1/k^2 gave you .96. That's the proportion which must under any circumstances lie between mean and 5 std dev from the mean. So the number is .96 n, where n is the number of elements in the sample. **
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RESPONSE --> ok
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01:55:06 query probl 13.3.48 mean length of stay 2.7 days, std dev 7.1 days.
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RESPONSE -->
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01:55:12 ** A sketch of a normal distribution will be a normal, or bell-shaped curve with its peak at the mean, dropping to about 60% of peak value at 1 std dev from the mean and to about 14% of peak value at 2 std dev from the mean. For a normal distribution with mean 2.7 and std dev. 7.1 one std dev from the mean occurs at 2.7 + 7.1 = 9.8 and at 2.7 7.1 = -4.4; two std dev from the mean occurs at 16.9 and -11.5. The corresponding normal curve cannot represent length of stay, since length of stay must not be less than zero. As a result we obtain a curve which tails off for large values of x, but whose area is concentrated mostly between 0 and 2.7. This curve is not symmetric like the bell curve but is very skewed, bunched up on one side of the mean 2.7 and more spread out for larger values. **
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RESPONSE --> ok
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01:56:05 **** Describe your sketch of the distribution of lengths of stay.
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RESPONSE --> The bell shape would go from 2.7 days at the lowest and 7.1 days at the highest
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01:56:18 2.7 is in the center with each number within 7.1 of the right or left of 2.7 and each additonal number on the left or right within 7.1 of each other. I see the curve as not being skewed.
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RESPONSE --> ok
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01:57:10 **** Is your distribution skewed? If so why, and if not why do you think it shouldn't be?
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RESPONSE --> yes, because of the deviation of 7.1
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01:57:18 STUDENT RESPONSE AND INSTRUCTOR COMMENT: I said no, but I'm not really sure how to determine this if I draw it myself, becasue I place all the numbers at an equal distance from each other. It just hit me while typing this that it must be skewed to the right because I can't really have negative days stay in the hospital. I think I have confused myself ** You didn't confuse yourself. That's exactly the point. You can't stay fewer than 0 days; since even 1 standard deviation is way below 0 the deviations must be primarily to the right of the mean. So the distribution must be skewed significantly to the right. GENERAL SUGGESTION: In general to understand the graphs of various distributions, try to understand in terms first of the bell-shaped curve with max height at the mean, dropping to about 60% height at a distance of 1 std dev from the mean and to about 14% at 2 std dev from the mean. Then understand that this distribution can be distorted, or skewed, as in this problem. This occurs when most of the distribution lies close to the mean on one side, with a smaller part of the distribution spread out further from the mean on the other. **
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RESPONSE --> ok
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01:57:38 017. `query 17
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RESPONSE --> ok
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The probability of ending up 6 blocks South is therefore prob = C(10,8) * (1/2)^8 * (1/2)^2 = 45 * (1/2)^10 = 45 / 1024, or about .044. **
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RESPONSE --> ok
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00:16:09 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
.................................................ÕCֈw
assignment #001 001. Only assignment: prelim asst
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12:11:49 `q001. Part 1 includes six activities. If you have completed an activity, just enter the answer 'completed'. This question is appearing in the Question box. The box to the right is the Answer box, where you will type in your answers to the questions posed here. To use this program you read a question, then enter your answer in the Answer box and click on Enter Answer. In your answers give what is requested, but don't go into excruciating detail. Try to give just enough that the instructor can tell that you understand an item. After entering an answer click on Next Question/Answer above the Question box. Do you understand these instructions?
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RESPONSE --> yes, I understand these instructions
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12:14:42 This program has created the folder c:\vhmthphy on your hard drive. Browse to that folder and locate the file whose name begins with SEND. The name of this file will also include your name, as you gave it to the program, and the file will show as a Text file. Never tamper with a SEND file in any way. It contains internal codes as if these codes are tampered with you won't get credit for the assignment. However you are welcome to copy this file to another location and view it, make changes, etc. Just be sure that when requested to do so you send the instructor the original, tamper-free file. State in the Answer box whether or not you have been able to locate the SEND file. Don't send the SEND file yet. Note that more questions/instructions remain in the q_a_prelim.
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RESPONSE --> I have located the SEND file and I understand not to send this file.
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12:15:36 `q002. Note that every time you click on Enter Answer the program writes your response to your SEND file. Even if the program disappears all the information you have entered with the Enter Answer button will remain in that file. This program never 'unwrites' anything. Even if this program crashes your information will still be there in the SEND file. Explain this in your own words.
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RESPONSE --> Whatever I type in the enter response section, stays there. The computer will not lose this information, even if it were to crash.
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12:17:56 Any time you do not receive a reply from the instructor by the end of the following day, you should resubmit your work using the Resubmit Form at http://www.vhcc.edu/dsmith/genInfo/. You have already seen that page, but take another look at that page and be sure you see the Submit Work form, the Resubmit Form and a number of other forms that will be explained later. Enter a sentence or two describing the related links you see at that location.
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RESPONSE --> http://www.vhcc.edu/dsmith/genInfo/ is the site for resubmitting my work.
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12:18:41 `q003. If you are working on a VHCC computer, it is probably set up in such a way as to return to its original configuration when it is rebooted. To avoid losing information it is suggested that you back up your work frequently, either by emailing yourself a copy or by using a key drive or other device. This is a good idea on any computer. Please indicate your understanding of this suggestion.
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RESPONSE --> I understand that if I am working on a VHCC computer that I need to email myself a copy of my work.
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|lOؐг assignment #001
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12:20:40 `q001. Part 1 includes six activities. If you have completed an activity, just enter the answer 'completed'. This question is appearing in the Question box. The box to the right is the Answer box, where you will type in your answers to the questions posed here. To use this program you read a question, then enter your answer in the Answer box and click on Enter Answer. In your answers give what is requested, but don't go into excruciating detail. Try to give just enough that the instructor can tell that you understand an item. After entering an answer click on Next Question/Answer above the Question box. Do you understand these instructions?
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RESPONSE --> I understand
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12:21:08 This program has created the folder c:\vhmthphy on your hard drive. Browse to that folder and locate the file whose name begins with SEND. The name of this file will also include your name, as you gave it to the program, and the file will show as a Text file. Never tamper with a SEND file in any way. It contains internal codes as if these codes are tampered with you won't get credit for the assignment. However you are welcome to copy this file to another location and view it, make changes, etc. Just be sure that when requested to do so you send the instructor the original, tamper-free file. State in the Answer box whether or not you have been able to locate the SEND file. Don't send the SEND file yet. Note that more questions/instructions remain in the q_a_prelim.
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RESPONSE --> already completed
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12:21:22 `q002. Note that every time you click on Enter Answer the program writes your response to your SEND file. Even if the program disappears all the information you have entered with the Enter Answer button will remain in that file. This program never 'unwrites' anything. Even if this program crashes your information will still be there in the SEND file. Explain this in your own words.
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RESPONSE --> already completed
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12:21:30 Any time you do not receive a reply from the instructor by the end of the following day, you should resubmit your work using the Resubmit Form at http://www.vhcc.edu/dsmith/genInfo/. You have already seen that page, but take another look at that page and be sure you see the Submit Work form, the Resubmit Form and a number of other forms that will be explained later. Enter a sentence or two describing the related links you see at that location.
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RESPONSE --> already completed
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12:21:45 `q003. If you are working on a VHCC computer, it is probably set up in such a way as to return to its original configuration when it is rebooted. To avoid losing information it is suggested that you back up your work frequently, either by emailing yourself a copy or by using a key drive or other device. This is a good idea on any computer. Please indicate your understanding of this suggestion.
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RESPONSE --> already completed
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12:21:49 Once more, locate the SEND file in your c:\vhmthphy folder, and open the file. Copy its contents to the clipboard (this is a common operation, but in case you don't know how, just use CTRL-A to highlight the contents of the file and CTRL-C to copy the contents to the clipboard). Then return to the form that instructed you to run this program, and paste the contents into the indicated box (just right-click in the box and select Paste). You may now click on the Quit button, or simply close the program.
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RESPONSE --> already completed
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y mӄWxBō~ Student Name: assignment #004 004. Liberal Arts Mathematics
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12:23:34 `q001. Consider the statement 'If that group of six-year-olds doesn't have adult supervision, they won't act in an orderly manner.' Under which of the following circumstances would everyone have to agree that the statement is false? The group does have supervision and they do act in an orderly manner. The group doesn't have supervision and they don't act in an orderly manner. The group doesn't have supervision and they do act in an orderly manner. The group does have supervision and they don't act in an orderly manner.
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RESPONSE --> The group doesn't have supervision and they don't act in an orderly manner.
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12:23:54 The statement says that if the group doesn't have supervision, they will not act in an orderly manner. So if they don't have supervision and yet do act in an orderly manner the statement is contradicted. If the group does have supervision, the statement cannot be contradicted because condition of the statement, that the group doesn't have supervision, does not hold. The statement has nothing to say about what happens if the group does have supervision. Of course if the group doesn't have supervision and doesn't act in orderly manner this is completely consistent with the statement. Therefore the only way to statement can be considered false is the group doesn't have supervision and does act in an overly manner. Note that what we know, or think we know, about childrens' behavior has nothing at all to do with the logic of the situation. We could analyze the logic of a statement like 'If the Moon is made of green cheese then most six-year-olds prefer collard greens to chocolate ice cream'. Anything we know about the composition of the Moon or the tastes of children has nothing to do with the fact that the only way this statement could be shown false would be for the Moon to be made of green cheese and most six-year-olds to prefer the ice cream.
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RESPONSE --> ok
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12:27:37 `q002. List the different orders in which the letters a, b and c could be arranged (examples are 'acb' and 'cba'). Explain how you know that your list contains every possible order.
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RESPONSE --> abc, acb bac, bca cba, cab If you only use each letter once in each list, this is the possibilities
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12:28:01 The only reliable way to get all possible orders is to have a system which we are sure the list every order without missing any. Perhaps the simplest way to construct all possible orders is to list then alphabetically. We start with abc. There is only one other order that starts with a, and it is obtained by switching the last two letters to get acb. The next alphabetical order must start with b. The first possible listing starting with b must follow b with a, leaving c for last. The orders therefore bac. The only other order starting with b is bca. The next order must start with c, which will be followed by a to give us cab. The next order is obtained by switching the last two letters to get cba. This exhausts all possibilities for combinations of the three letters a, b and c. Our combinations are, in alphabetical order, abc, acb, bac, bca, cab, cba.
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RESPONSE --> ok
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12:31:04 `q003. One collection consists of the letters a, c, d and f. Another collection consists of the letters a, b, d and g. List the letters common to both collections. List the letters which appear in at least one of the collections. List the letters in the first half of the alphabet which do not appear in either of the collections.
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RESPONSE --> Letters common to both: a, d Letters in at least one collection: a, b, c, d, f, g Letters in the first half of the alphabet which do not appear in either collection: e
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12:31:46 To letters a and d each appear in both collections. No other letter does. The letters a, c, d, and f appear in the first collection, so they all in at least one of the collections. In addition to letters b and g appear in the second collection. Therefore letters a, b, c, d, f and g all appear in at least one of the collections. We consider the letters in the first half of the alphabet, in alphabetical order. a, b, c and d all appear in at least one of the collections, but the letter e does not. The letters f and g also appear in at least one of the collections, but none of the other letters of the alphabet do. The first half of the alphabet ends at m, so the list of letters in the first half of the alphabet which do not occur in at least one of the collections is e, h, i, j, k, l, m.
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RESPONSE --> ok
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12:34:30 `q004. Give the next element in each of the following patterns and explain how you obtained each: 2, 3, 5, 8, 12, ... 3, 6, 12, 24, ... 1, 3, 4, 7, 11, 18, ...
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RESPONSE --> 2, 3, 5, 8, 12, 17 3, 6, 12, 24, 48 1, 3, 4, 7, 11, 18, 29
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12:34:43 The pattern of the sequence 2, 3, 5, 8, 12, ... can be seen by subtracting each number from its successor. 3-2 = 1, 5-3 = 2, 8-5 = 3, 12-8 = 4. The sequence of differences is therefore 1, 2, 3, 4, ... . The next difference will be 5, indicating that the next number must be 12 + 5 = 17. The pattern of the sequence 3, 6, 12, 24, ... can be discovered by dividing each number into its successor. We obtain 6/3 = 2, 12/6 = 2, 24/12 = 2. This shows us that we are doubling each number to get the next. It follows that the next number in the sequence will be the double of 24, or 48. The pattern of the sequence 1, 3, 4, 7, 11, 18, ... is a little obvious. Starting with the third number in the sequence, each number is the sum of the two numbers proceeding. That is, 1 + 3 = 4, 3 + 4 = 7, 4 + 7 = 11, and 7 + 11 = 18. It follows that the next member should be 11 + 18 = 29.
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RESPONSE --> ok
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12:38:21 `q005. The number 18 can be 'broken down' into the product 9 * 2, which can then be broken down into the product 3 * 3 * 2, which cannot be broken down any further . Alternatively 18 could be broken down into 6 * 3, which can then be broken down into 2 * 3 * 3. Show how the numbers 28 and 34 can be broken down until they can't be broken down any further. Show that there at least two different ways to break down 28, but that when the breakdown is complete both ways end up giving you the same numbers.
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RESPONSE --> 28: 7*4, 7*2*2 34: 17*2
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12:38:33 A good system is to begin by attempting to divide the smallest possible number into the given number. In the case of 34 we see that the number can be divided by 2 give 34 = 2 * 17. It is clear that the factor 2 cannot be further broken down, and is easy to see that 17 cannot be further broken down. So the complete breakdown of 34 is 2 * 17. To breakdown 28 we can again divide by 2 to get 28 = 2 * 14. The number 2 cannot be further broken down, but 14 can be divided by 2 to give 14 = 2 * 7, which cannot be further broken down. Thus we have 28 = 2 * 2 * 7. The number 28 could also the broken down initially into 4 * 7. The 4 can be further broken down into 2 * 2, so again we get 28 = 2 * 2 * 7. It turns out that the breakdown of a given number always ends up with exactly same numbers, no matter what the initial breakdown.
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RESPONSE --> ok
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12:40:42 `q006. Give the average of the numbers in the following list: 3, 4, 6, 6, 7, 7, 9. By how much does each number differ from the average?
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RESPONSE --> The average is 6. The differences from this average are 3, 2, 0, 0, 1, 1, 3
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12:41:04 To average least 7 numbers we add them in divide by 7. We get a total of 3 + 4 + 6 + 6 + 7 + 7 + 9 = 42, which we then divide by 7 to get the average 42 / 7 = 6. We see that 3 differs from the average of 6 by 3, 4 differs from the average of 6 by 2, 6 differs from the average of 6 by 0, 7 differs from the average of 6 by 1, and 9 differs from the average of 6 by 3. A common error is to write the entire sequence of calculations on one line, as 3 + 4 + 6 + 6 + 7 + 7 + 9 = 42 / 7 = 6. This is a really terrible habit. The = sign indicates equality, and if one thing is equal to another, and this other today third thing, then the first thing must be equal to the third thing. This would mean that 3 + 4 + 6 + 6 + 7 + 7 + 9 would have to be equal to 6. This is clearly not the case. It is a serious error to use the = sign for anything but equality, and it should certainly not be used to indicate a sequence of calculations.
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RESPONSE --> ok
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12:42:33 `q007. Which of the following list of numbers is more spread out, 7, 8, 10, 10, 11, 13 or 894, 897, 902, 908, 910, 912? On what basis did you justify your answer?
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RESPONSE --> The larger numbers are more spread out than the smaller numbers. They differ less.
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12:42:52 The first set of numbers ranges from 7 to 13, a difference of only 6. The second set ranges from 894 to 912, a difference of 18. So it appears pretty clear that the second set has more variation the first. We might also look at the spacing between numbers, which in the first set is 1, 2, 0, 1, 2 and in the second set is 3, 5, 6, 2, 2. The spacing in the second set is clearly greater than the spacing in the first. There are other more sophisticated measures of the spread of a distribution of numbers, which you may encounter in your course.
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RESPONSE --> ok
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12:46:31 `q008. 12 is 9 more than 3 and also 4 times 3. We therefore say that 12 differs from 3 by 9, and that the ratio of 12 to 3 is 4. What is the ratio of 36 to 4 and by how much does 36 differ from 4? If 288 is in the same ratio to a certain number as 36 is to 4, what is that number?
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RESPONSE --> Ratio of 36 to 4 is 9. 36 differs from 4 by 32. Ratio of 288 to 36 is 32.
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12:46:58 Just as the ratio of 12 to 3 is 12 / 3 = 4, the ratio of 36 to 4 is 36 / 4 = 9. 36 differs from 4 by 36 - 4 = 32. Since the ratio of 36 to 4 is 9, the number 288 will be in the same ratio to a number which is 1/9 as great, or 288 / 9 = 32. Putting this another way, the question asks for a 'certain number', and 288 is in the same ratio to that number as 36 to 4. 36 is 9 times as great as 4, so 288 is 9 times as great as the desired number. The desired number is therefore 288/9 = 32.
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RESPONSE --> ok
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12:48:44 `q009. A triangle has sides 3, 4 and 5. Another triangle has the identical shape of the first but is larger. Its shorter sides are 12 and 16. What is the length of its longest side?
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RESPONSE --> the longest side would be 20
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12:48:56 ** You need to first see that that each side of the larger triangle is 4 times the length of the corresponding side of the smaller. This can be seen in many ways, one of the most reliable is to check out the short-side ratios, which are 12/3 = 4 and 16/4 = 4. Since we have a 4-to-1 ratio for each set of corresponding sides, the side of the larger triangle that corresponds to the side of length 5 is 4 * 5 = 20. **
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RESPONSE --> ok
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nEcwڌ撏 Student Name: assignment #001 001. typewriter notation
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12:52:53 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> x - 2 / x + 4 and (x - 2) / (x + 4) When you have parenthesis, you must get that answer first
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12:53:00 The order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction. It makes a big difference whether you subtract the 2 from the 2 or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract: 2 - 2 / 2 + 4 = 2 - 1 + 4 (do multiplications and divisions before additions and subtractions) = 5 (add and subtract in indicated order) If there are parentheses you evaluate the grouped expressions first: (x - 2) / (x - 4) = (2 - 2) / ( 4 - 2) = 0 / 2 = 0.
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RESPONSE --> ok
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12:58:26 `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2. Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.
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RESPONSE --> The first problem can be completed from left to right but the second has parenthesis, so that part must be completed first. x=2 2*2*2=8 8+4=12 2+4=8 2*2*2*2*2*2*2*2*2=512
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12:58:36 2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4. 2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power. If x = 2, then 2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8. and 2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.
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RESPONSE --> ok
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13:06:27 `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?
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RESPONSE --> x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x x = 2 2-3/ [ (2*2-5)^2*3*2+1] -2+7*2 2-3/ [(-1)^2*7]-2+14 2-3/[-1]-2+14 2-3/11 -1/11 -0.09
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13:06:44 The numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3. If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way. The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ] Evaluating the expression for x = 2: - 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 = 2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses 2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses 2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term; 2 - 3 / 7 - 2 + 14 = evaluate in brackets 13 4/7 or 95/7 or about 13.57 add and subtract in order. The details of the calculation 2 - 3 / 7 - 2 + 14: Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have 2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.
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RESPONSE --> ok
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13:11:57 `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
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RESPONSE --> x=4 (4-5)= -1 2x-1= 8-1=7 -1^7= 1 3/x-2= 3/4-2= 3/2 1+3/2= 4/2= 2
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13:12:56 We get (4-5)^2 * 4 - 1 + 3 / 1 - 4 = (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses = 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4 = 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get = 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4). COMMON ERROR: (4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 = -1 ^ 2*4 - 1 + 3 / 4-2 = -1 ^ 8 -1 + 3 / 4 - 2. INSTRUCTOR COMMENTS: There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first. Exponentiation precedes multiplication. Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4). Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power. -1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.
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RESPONSE --> ok
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Ӆx|LK̊ Student Name: assignment #002 002. Describing Graphs
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13:24:59 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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RESPONSE --> y=3x-4 y=3*-3-4 y=-9-4 y= -13 y=3x-4 y= 3*-2-4 y=-6-4 y= -10 y=3x-4 y=3*-1-4 y=-3-4 y= -7 y=3x-4 y=3*0-4 y=0-4 y= -4 y=3x-4 y=3*1-4 y=3-4 y= -1 y=3x-4 y=3*2-4 y=6-4 y= 2 y=3x-4 y=3*3-4 y=9-4 y= 5 (-3, -13) (-2, -10) (-1, -7) (0, -4) (1, -1) (2, 2) (3, 5)
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13:25:23 The graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.
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RESPONSE --> ok
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13:28:04 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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RESPONSE --> The steepness increases from left to right
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13:28:45 The graph forms a straight line with no change in steepness.
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RESPONSE --> ok
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13:30:11 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> 3/3, The slope rises 3 for every 3 numbers
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13:30:18 Between any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.
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RESPONSE --> ok ....
..... CpF`
assignment #003 cʓ齯̀q} li Liberal Arts Mathematics II 10-10-2006
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15:23:15 Query 11.3.20 5 prizes among 25 students
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RESPONSE --> ok
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15:24:36 ** There are 25 students available so there are 25 choices for the first student. On the second choice there are 24 students left so there would be 24 possibilities. Similarly on the third, fourth and fifth selections there would be 23, 22 and 21 choices. The result, by the Fundamental Counting Principle, is 25 * 24 * 23 * 22 * 21 choices. 25 * 24 * 23 * 22 * 21 = 25 ! / ( (25 - 5) !) since 25 ! / ( (25 - 5) !) = 25 ! / (20! * 5!) = 25 * 24 * 23 * 22 * 21 This is P(25, 5). We use permutations because order matters when there are 5 different prizes. **
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RESPONSE --> ok, sorry, I typed ok, too fast
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15:25:26 Is repetition allowed in this situation?
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RESPONSE --> no
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15:25:41 ** GOOD STUDENT ANSWER: no repetition is allowed because there are 5 different prizes, and you can't give the same one to two people **
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RESPONSE --> exactly
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15:29:27 Query 11.3.30 3-letter monogram all letters different
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RESPONSE --> N={A,B,C} {A,B,C}
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15:30:24 ** You are choosing 3 different letters, and since the monogram will be different if you change the order we can say that order definitely applies. If there is no restriction on any letter, other than the restriction of no repetitions, then there are 26 choices for the first letter, 25 for the second, 24 for the third so by the Fundamental Counting Principle there are 26 * 25 * 24 ordered choices. We can write this as P(26, 3), the number of possible permutations of 3 objects chosen without replacement from 26. P(26,3) = 26!/(26-3) ! = 26 * 25 * 24, in agreement with the previous expression. However the third initial must be the same as Judy's, which is `z'. Thus, since there can be no repetitions, there are only 25 possibilities for the first letter (can't be `z') and 24 for the second (can't be `z', can't be the first). So there are only 25 * 24 = 600 possibilities. **
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RESPONSE --> OK
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15:42:29 Query 11.3.42 divide 25 students into groups of 3,4,5,6,7. In how many ways can the students be grouped?
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RESPONSE --> C(N,R)=P(N,R)/R!=N!/R!(N-R)! C(25,3)=P(25,3)/3!=25!/3!(25-3)! 25= 25/66 0.378
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15:43:09 ** If we make the group of 3 first there are C(25, 3) possible choices. If we make the group of 4 next there are 22 students left from whom to choose so there are C(22, 4) possible choices. If we make the group of 5 next there are 18 students left from whom to choose so there are C(18, 5) possible choices. If we make the group of 6 next there are 13 students left from whom to choose so there are C(13, 6) possible choices. If we make the group of 7 next there are 7 students left from whom to choose so there are C(7, 7) possible choices. The Fundamental Counting Principle tells you that you have to multiply the number of ways of obtaining the first group by the number of ways of obtaining the second group by the number of ways of obtaining the 3rd group by the number of ways of obtaining the fourth group by the number of ways of obtaining the fifth group: C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7). Note that we could have chosen the groups in a different order, perhaps with the group of 7 first, the group of 6 second, etc.. The same reasoning would tell us that there are now C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) ways to do this. If the two expressions C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7) and C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) are both written down and simplified can get them both into the same form 25! / [ 3 ! * 4 ! * 5 ! * 6 ! * 7 ! ], which can then be further simplified by cancellation and then multiplied. **
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RESPONSE --> OK, I UNDERSTAND WHAT YOU WERE LOOKING FOR NOW
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15:50:08 Query 11.3.60 C(n,0)What is the value of C(n,0)?What is the value of C(8,0)?
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RESPONSE --> C(n,0)=0, N!=N-1 C(8,0)=40,319!
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15:51:39 ** This is equivalent to choosing 0 objects from n objects. No matter what n is, there is one way to do this, which is to choose nothing. As another example there are C(4,2) = 6 ways in which to obtain 2 Heads on four flips of a coin, C(4,3) = 4 ways to obtain 3 Heads, C(4,4) = 1 way to obtain 4 Heads. Obtaining 4 Heads is the same as obtaining 0 Tails, and of course C(4,0) tells you how many ways that are to obtain 0 Tails. So C(4,0) must be 1. The formula also gives us this: C(n, 0) = n ! / [ (n - 0) ! * 0 ! ] = n ! / ( n ! * 1) = 1. **
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RESPONSE --> OK, I'M CONFUSED. THIS PROBLEM DOESN'T MATCH UP WITH THE #60 IN THE BOOK
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秇i~~^}jD assignment #004 cʓ齯̀q} li Liberal Arts Mathematics II 10-10-2006
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15:54:59 Query 11.4.6 Find C(9,6) on Pascal's triangle.
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RESPONSE --> C(9,6) is located on row 9 of the triangle
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15:55:33 ** You need to go to the n=9 row, the r=6 position, which is the 10th row and the 7th number in the row. Looks like you went to the 6th position. Note that C(9,6) = 9! / ( 6! (6-3)! ) = 9*8*7 / (3*2*1) = 3 * 4 * 7 = 84 does agree with the number in the n = 9 row and the r = 6 position. Note also that since the first row is row 0 and the first element in every row is element 0, the n = 9 row is the 10th row, and the r = 6 position is the 7th number from the left. **
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RESPONSE --> I understand to start with row 0.
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15:59:52 Query 11.4.18 clueless check of four of nine possible classroomsHow many of the possible selections will fail to locate the classroom?
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RESPONSE --> 70 selections will fail
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16:00:20 ** There are C(9,4) possible combinations of four classrooms. There are 8 'incorrect' classrooms, so there are C(8, 4) ways for the check to yield a 'wrong' classroom. C(9,4) = 126 and C(8,4) = 70. Note that the chance of ending up in the right classroom is 56 / 126, a little less than 50-50, with 56 of the 126 possible ways being successful. **
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RESPONSE --> ok
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16:02:05 Query 11.4.30 What sequence by totaling diagonals of Pascal's Triangle?
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RESPONSE --> random walks
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16:04:33 ** The numbers 1, 1, 2, 3, 5, 8, 13, 21, ... form what is called the Fibonacci Sequence, which seems to occur in all sorts of unexpected places. The sums of the diagonals are all Fibonacci numbers. **
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RESPONSE --> This was something discussed in chapter 5, not chapter 11
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16:17:52 Query 11.4.42 (x+y)^8
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RESPONSE --> x^8+8x^7y+28x^6y^2+56x^5y^3+70x^4y^4+56x^3y^5+28x^2y^6+8xy^7+y^8
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16:18:30 **(x + y ) ^ 8 = x^8 + C(8,1) x^7 y + C(8,2) x^6 y^2 + C(8,3) x^5 y^3 + C(8,4) x^4 y^4 + C(8,5) x^3 y^5 + C(8,6) x^2 y^6 + C(8,7) x^7 y + y^8 = x^8 + 8 x^7 y + 28 x^6 y^2 + 56 x^5 y^3 + 70 x^4 y^4 + 56 x^3 y^5 + 28 x^2 y^6 + 8 x y^7 + y^8. **
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RESPONSE --> ok
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16:19:02 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> No comments
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16:20:29 ** STUDENT COMMENT: I was ok with this assignment until I got to problem 11.4 - 42 I do not understand the reasoning behind the following problems. 11.4 - 45: (2a + 5b)^4 = The binomial expansion is listed in the answer section, but I do not understand how they got there. INSTRUCTOR RESPONSE: Here is the solution for (2a + 5b) ^ 5. The application of the Binomial Theorem is clearer for the 5th power than the 4th; if you understand this you'll get the pattern for the 4th power. The answer is found from C(5,5) * (2a)^2 * (5b)^0 + C(5, 4) * (2a)^4 * (5b)^1 + C(5, 2) * (2a)^3 * (5b)^2 + . etc., following the pattern of the binomial expansion formula. Expanding the powers of 2a and 5b we get C(5, 5) * 32 a^5 + C(5, 4) * 16 a^4 * 5b + C(5, 3) * 8 a^3 * 25 b^2 + etc., which is equal to 1 * 32 a^5 + 5 * 16 a^4 * 5b + 10 * 8 a^3 * 25 b^2 + etc., or finally to 32 a^5 + 80 a^4 + 2000 a^3 + etc.. ANOTHER QUESTION: 11.4 -50 and 51 The rth or general term of the binomial expansion for (x = y)^n and (x + y)^14;5th term. INSTRUCTOR RESPONSE: The rth term of (x+y)^n will be C(n, r) * x^r * y^(n-r). You are choosing x from r of the binomials and y from the remaining n - r binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) . (x+y), where it is understood that we have (x+y) written n times. The 5th term of (x+y)^14 requires that you choose x from 5 of the binomials and y from the other 14-5 = 9 binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y). There are C(14, 5) ways to do this, and the result for every one of these ways is x^5 * y^14. So the 5th term is C(14,5) x^5 y^9. **
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RESPONSE --> ok, I finally got the hang of this assignment toward the end, only the last couple actually. I had a really hard time with this whole assignment. The book doesn't do a very good job with explaining these problems.
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{cϙys assignment #005 cʓ齯̀q} li Liberal Arts Mathematics II 10-10-2006
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17:33:08 Query 11.5.10 ways to get different number on each of two fair dice In how many ways can you get a different number on each of two fair dice? How would you obtain the answer to this question if you were marooned on a desert island with no book and no table?
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RESPONSE --> 72 ways, 36 chances for each di
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17:33:37 ** On two fair dice you have 6 possible outcomes on the first and 6 on the second. By the Fundamental Counting Principle there are therefore 6 * 6 = 36 possible outcomes. We can list these outcomes in the form of ordered pairs: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) Of these 36 outcomes there are six that have the same number on both dice. It follows that the remaining 3 - 6 = 30 have different numbers. So there are 30 ways to get different numbers on the two dice. Note that your chance of getting different numbers is therefore 30 / 36 = 5/6 = .8333... .**
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RESPONSE --> ok, I understand not to double and why
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17:34:52 Query 11.5.12 bridge hands more than one suit How many bridge hands contain more than one suit?
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RESPONSE --> 36
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17:35:53 ** There are 13 cards in a bridge hand. The number of possible bridge hands is therefore C(52, 13). There are 13 cards of each suit. The number of possible bridge hands with all cards in a given suit is therefore C(13, 13) = 1 (common sense is that there is only one way to get all 13 cards in a given suit, which is to get all the cards there are in that suit). Since there are 4 suits there are 4 * C(13, 13) = 4 * 1 = 4 possible one-suit hands. Of the number of hands having more than one suit is C(52, 13) - 4. **
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RESPONSE --> ok
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17:48:15 11.5.20 # subsets of 12-elt set with from 3 to 9 elts? How many subsets contain from three to nine elements and how did you obtain your answer (answer in detail)?
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RESPONSE --> 8^2=64 64-1=63 8 restaraunts total, only 2 serve seafood. Only 1of the selected restaraunts will be serving seafood.
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17:50:12 ** You need the number of subsets with 3 elements, with 4 elements, etc.. You will then add these numbers to get the total number of 3-, 4-, 5-, ., 9-element subsets. Start with a 3-e.ement subset. In a 12-element set, how many subsets have exactly three elements? You answer this by asking how many possibilities there are for the first element, then how many for the second, then how many for the third. You can choose the first element from the entire set of 12, so you have 12 choices. You have 11 elements from which to choose the second, so there are 11 choices. You then have 10 elements left from which to choose the third. So there are 12 * 11 * 10 ways to choose the elements. However, the order of a set doesn't matter. 3 elements could be ordered in 3! different ways, so there are 12 * 11 * 10 / 3! ways to choose different 3-element sets. This is equal to C(12,3). So there are C(12, 3) 3-elements subsets of a set of 12 elements. Reasoning similarly we find that there are C(12,4) ways to choose a 4-element subset. C(12,5) ways to choose a 5-element subset. C(12,6) ways to choose a 6-element subset. C(12,7) ways to choose a 7-element subset. C(12,8) ways to choose a 8-element subset. C(12,9) ways to choose a 9-element subset. We see that there are C(12,3) + C(12,4) + C(12,5) + C(12,6) + C(12,7) + C(12,8) + C(12,9) possible subsets with 3, 4, 5, 6, 7, 8 or 9 elements. Alternatively you can figure out how many sets have fewer than 3 or more than 9 elements. There are C(12, 0) + C(12, 1) + C(12, 2) = 1 + 12 + 66 = 79 sets with fewer than 3 elements, and C(12, 10) + C(12, 11) + C(12, 12) = 66 + 12 + 1 = 79 sets with more than 3. Since there are 2^12 = 4096 possible subsets of a 12-element set there are 4096 - 79 - 79 = 3938 sets with between 3 and 9 elements. **
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RESPONSE --> ok
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18:04:03 11.5.30 10200 ways to get a straight Verify that there are in fact 10200 ways to get a straight in a 5-card hand.
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RESPONSE --> C(52,5) 5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+52=1368 2,598,960-C(52,5)=2,597,592
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18:04:44 ** There are 9 choices for the denomination of the lowest card in a straight, which gives 36 cards that could be the low card. However but if aces can be high or low there are 40. There are then four choices for the next-higher card, four for the next after that, etc., giving 40*4*4*4*4 possibilities. **STUDENT COMMENT: I don 't understand this one . Idon't see where you get the 9 from. INSTRUCTOR RESPONSE: Cards run from 2 through 10, then the four face cards, then the ace. You need five consecutive cards to make a straight. The highest possible straight is therefore 10, Jack, Queen, King and Ace. The lowest is 2, 3, 4, 5, 6. The lowest card of the straight can be any number from 2 through 10. That is 9 possibilities. In some games the ace can be counted as the low card, below the 2, as well as the high card. In that case there would be one more possibility for a straight, which could then consists of denominations 1, 2, 3, 4, 5. *&*&
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RESPONSE --> ok, I don't understand
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18:17:39 11.5.36 3-digit #'s from {0, 1, ..., 6}; how many mult of 25?
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RESPONSE --> 3- digit #'s from {0,1,2,3,4,5,6} C(25,1)=25 C(25,2)=300
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18:20:14 ** A 3-digit number from the set has six choices for the first digit (can't start with 0) and 7 choices for each remaining digit. That makes 6 * 7 * 7 = 294 possibilities. A multiple of 25 is any number that ends with 00, 25, 50 or 75. SInce 7 isn't in the set you can't have 75, so there are three possibilities for the last two digits. There are six possible first digits, so from this set there are 6 * 3 = 18 possible 3-digit numbers which are multiples of 25. A listing would include 100, 125, 150, 200, 225, 250, 300, 325, 350, 400, 425, 450, 500, 525, 550, 600, 625, 650. Combinations aren't appropriate for two reasons. In the first place the uniformity criterion is not satisfied because different digits have different criteria (i.e., the first digit cannot be zero). In the second place we are not choosing object without replacement. The fundamental counting principle is the key here. STUDENT SOLUTION AND INSTRUCTOR RESPONSE: All I can come up with is C(7,2)=21. & choices of #s and the # must end in 0 or 5 making it 2 of the 7 choices INSTRUCTOR RESPONSE: Right reasoning on the individual coices but you're not choosing just any 3 of the 7 numbers (uniformity criterion isn't satisfied--second number has different criterion than first--so you wouldn't use permutations or combinations) and order does matter in any case so you wouldn't use combinations. You have 7 choices for the first and 2 for the second number so there are 7 * 2 = 14 multiples of 5. **
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RESPONSE --> I understand not to start with 0. Ok, I think I might understand now, after reading through your reasoning
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18:39:58 Query 11.5.48 # 3-digit counting #'s without digits 2,5,7,8?
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RESPONSE --> If you figure up 3-digit numbers that do not contain the digits 2,5,7,8, there are only 200 numbers left.
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18:40:36 ** there are 5 possible first digits (1, 3, 4, 6, or 9) and 6 possibilities for each of the last two digits. This gives you a total of 5 * 6 * 6 = 180 possibilities. **
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RESPONSE --> Ok, I still don't understand
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18:41:02 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I really had trouble with this section. I think I will try to get some assistance.
....... 㼯izUw[_Ȋjqy
assignment #006 cʓ齯̀q} li Liberal Arts Mathematics II 10-11-2006
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19:00:08 Query 12.1.6 8 girls 5 boys What is the probability that the first chosen is a girl?
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RESPONSE --> 13 people all together, 8 girls 8/13=0.62
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19:00:23 ** Assuming the choice is completely random there are 13 possible choices, 8 of which are female so we have P(female) = 8 / 13 = .6154, approx. **
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RESPONSE --> ok, I understand. I just rounded up
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19:05:30 Query 12.1.12 3 fair coins: Probability and odds of 3 Heads.
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RESPONSE --> 1/2 for each coin= 0.50 3/1=1/1 are the odds
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19:06:26 ** There are 8 equally likely possible outcomes when flipping 3 fair coins. You can list them: hhh, hht, hth, htt, thh, tht, tth, ttt. Or you can use the fact that there are 2 possibilities on each flip, therefore 2*2*2 = 2^3 = 8 possible outcomes. Only one of these outcomes, hhh, consists of 3 heads. The probability is therefore P(3 heads) = # of outcomes favorable/total number of possible outcomes = 1 / 8. The odds in favor of three heads are Odds ( 3 heads ) = # favorable to # unfavorable = 1 to 7. **
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RESPONSE --> ok
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19:12:14 Query 12.1.20 P(pink) from two pink parents (Rr and Rr) What is the probability of a pink offspring.
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RESPONSE --> If both parents are Pink, there should be a 100% chance of a pink offspring. However, Do the parents not still possess the red and white passed from their parents? If so, that is still s recessive trait and the chance is lowered to 25% to 75% depending on how you look at it.
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19:12:39 ** The genes R and r stand for the red and white genes. A pink offspring is either Rr or rR. RR will be red, rr white. R r R RR Rr r rR rr shows that {RR, Rr, rR, rr} is the set of equally likely outcomes. We season two of the four possible outcomes, rR and Rr, will be pink. So the probability of pink offspring is 2/4 = 1 / 2. **
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RESPONSE --> Ok, you are going by dominence. I understand
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19:15:42 Query 12.1.33 cystic fibrosis in 1 of 2K cauc, 1 in 250k noncauc What is the empirical probability, to 6 places, that a randomly chosen non-Caucasian newborn will have cystic fibrosis?
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RESPONSE --> 1/500=0.002 because this is based on the information given in problem 33. For CF, it is 0.000004
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19:15:48 ** There is 1 chance in 250,000 so the probability is 1 / 250,000 = 4 * 10^-6, or .000004. **
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RESPONSE --> ok
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19:18:07 12.1.40 Cc genes carrier, cc has disease; 2 carriers first child has disease **** What is the probability that the first child has the disease?
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RESPONSE --> If you write this out in table form, 1/4=0.25. This is the value for cc. There is also a cC so wouldn't they have it in some form as well, making it 0.50?
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19:18:37 ** If cc has the disease, then the probability that the first child will have the disease is 1/4. **
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RESPONSE --> ok, I see, it is only the obvious and I shouldn't look further.
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19:19:34 What is the sample space for this problem?
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RESPONSE --> I'm not sure what you are asking
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19:20:43 ** The sample space is {CC, Cc. cC, cc}. **
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RESPONSE --> oh, ok, that is what I meant by the table. Mine was the same and is below. C c C CC Cc c cC cc
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19:21:42 12.1.60 36 in class, 3 chosen **** What is the probability that the choice will be the given three people in any order?
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RESPONSE --> 3/36=0.083
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19:22:04 ** There are P(36,3) possible ordered choices of 3 people out of the 36. P(36,3) = 36! / (36-3)! = 36! / 33! = (36*35*34*33*32*31*...*1) / (33*32*31*...*1) = 36*35*34=40,000 or so. The probability of any given choice is therefore 1 / P(36,3) = 1/40,000 = .000025, approx.. For any given set of three people there are six possible orders in which they can be chosen. So the probability of the three given people, in any order, is 6 * probability of a given order = 6 / P(36,3) = 6/40,000 = .00015. Alternatively we can say that we are choosing 3 of 36 people without regard for order, so there are C(36,3) possibilities and the probability of any one of them is 1 / C(36,3). **
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RESPONSE --> ok
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19:26:02 Query 12.1.75 digits 1, 2, ..., 5 rand arranged; prob even, prob digits 1 and 5 even What is the probability that the resulting number is even and how did you obtain your answer?
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RESPONSE --> 2 and 4 are the even numbers in this problem 1+2+3+4+5=15 15/75=0.2
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19:26:40 ** The number will be even if it ends in 2 or 4. There are 5 possible ending numbers. So 2 of the 5 possible ending numbers are even and the probability of an even number is 2/5. We analyze in two ways the number of ways to choose a number with digits 1 and 5 even. First way: There are 5! = 120 possible arrangements of the 5 digits. There are only two possible even digits, from which we will choose digit 1 and digit 5. The order of our choice certainly matters, since a different choice will give us a different 5-digit number. So we are choosing 2 digits from a set of 2 digits, where order matters. We therefore have P(2, 2) = 2*1 / 0! = 2 ways to choose these digits. The remaining 3 digits will comprise digits 2, 3 and 4. We are therefore choosing 3 digits from a set of 3, in order. There are P(3, 3) = 3*2*1/0! = 6 ways to do so. To obtain our number we can choose digits 1 and 5, then digits 2, 3 and 4. There are P(2, 2) * P(3, 3) = 2 * 6 = 12 ways to do this. So the probability that digits 1 and 5 are even is 12 / 120 = 1/10. Second way: A simpler solution looks at just the possibilities for digits 1 and 5. There are P(2, 2) = 2 choices for which these digits are even, and P(5, 2) = 20 total choices for these two digits. The probability that both will be even is therefore 2/20 = 1/10, the same as before. **
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RESPONSE --> ok
...................................................... 㼯izUw[_Ȋjqy
assignment #006 cʓ齯̀q} li Liberal Arts Mathematics II 10-11-2006
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19:00:08 Query 12.1.6 8 girls 5 boys What is the probability that the first chosen is a girl?
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RESPONSE --> 13 people all together, 8 girls 8/13=0.62
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19:00:23 ** Assuming the choice is completely random there are 13 possible choices, 8 of which are female so we have P(female) = 8 / 13 = .6154, approx. **
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RESPONSE --> ok, I understand. I just rounded up
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19:05:30 Query 12.1.12 3 fair coins: Probability and odds of 3 Heads.
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RESPONSE --> 1/2 for each coin= 0.50 3/1=1/1 are the odds
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19:06:26 ** There are 8 equally likely possible outcomes when flipping 3 fair coins. You can list them: hhh, hht, hth, htt, thh, tht, tth, ttt. Or you can use the fact that there are 2 possibilities on each flip, therefore 2*2*2 = 2^3 = 8 possible outcomes. Only one of these outcomes, hhh, consists of 3 heads. The probability is therefore P(3 heads) = # of outcomes favorable/total number of possible outcomes = 1 / 8. The odds in favor of three heads are Odds ( 3 heads ) = # favorable to # unfavorable = 1 to 7. **
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RESPONSE --> ok
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19:12:14 Query 12.1.20 P(pink) from two pink parents (Rr and Rr) What is the probability of a pink offspring.
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RESPONSE --> If both parents are Pink, there should be a 100% chance of a pink offspring. However, Do the parents not still possess the red and white passed from their parents? If so, that is still s recessive trait and the chance is lowered to 25% to 75% depending on how you look at it.
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19:12:39 ** The genes R and r stand for the red and white genes. A pink offspring is either Rr or rR. RR will be red, rr white. R r R RR Rr r rR rr shows that {RR, Rr, rR, rr} is the set of equally likely outcomes. We season two of the four possible outcomes, rR and Rr, will be pink. So the probability of pink offspring is 2/4 = 1 / 2. **
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RESPONSE --> Ok, you are going by dominence. I understand
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19:15:42 Query 12.1.33 cystic fibrosis in 1 of 2K cauc, 1 in 250k noncauc What is the empirical probability, to 6 places, that a randomly chosen non-Caucasian newborn will have cystic fibrosis?
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RESPONSE --> 1/500=0.002 because this is based on the information given in problem 33. For CF, it is 0.000004
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19:15:48 ** There is 1 chance in 250,000 so the probability is 1 / 250,000 = 4 * 10^-6, or .000004. **
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RESPONSE --> ok
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19:18:07 12.1.40 Cc genes carrier, cc has disease; 2 carriers first child has disease **** What is the probability that the first child has the disease?
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RESPONSE --> If you write this out in table form, 1/4=0.25. This is the value for cc. There is also a cC so wouldn't they have it in some form as well, making it 0.50?
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19:18:37 ** If cc has the disease, then the probability that the first child will have the disease is 1/4. **
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RESPONSE --> ok, I see, it is only the obvious and I shouldn't look further.
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19:19:34 What is the sample space for this problem?
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RESPONSE --> I'm not sure what you are asking
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19:20:43 ** The sample space is {CC, Cc. cC, cc}. **
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RESPONSE --> oh, ok, that is what I meant by the table. Mine was the same and is below. C c C CC Cc c cC cc
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19:21:42 12.1.60 36 in class, 3 chosen **** What is the probability that the choice will be the given three people in any order?
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RESPONSE --> 3/36=0.083
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19:22:04 ** There are P(36,3) possible ordered choices of 3 people out of the 36. P(36,3) = 36! / (36-3)! = 36! / 33! = (36*35*34*33*32*31*...*1) / (33*32*31*...*1) = 36*35*34=40,000 or so. The probability of any given choice is therefore 1 / P(36,3) = 1/40,000 = .000025, approx.. For any given set of three people there are six possible orders in which they can be chosen. So the probability of the three given people, in any order, is 6 * probability of a given order = 6 / P(36,3) = 6/40,000 = .00015. Alternatively we can say that we are choosing 3 of 36 people without regard for order, so there are C(36,3) possibilities and the probability of any one of them is 1 / C(36,3). **
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RESPONSE --> ok
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19:26:02 Query 12.1.75 digits 1, 2, ..., 5 rand arranged; prob even, prob digits 1 and 5 even What is the probability that the resulting number is even and how did you obtain your answer?
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RESPONSE --> 2 and 4 are the even numbers in this problem 1+2+3+4+5=15 15/75=0.2
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19:26:40 ** The number will be even if it ends in 2 or 4. There are 5 possible ending numbers. So 2 of the 5 possible ending numbers are even and the probability of an even number is 2/5. We analyze in two ways the number of ways to choose a number with digits 1 and 5 even. First way: There are 5! = 120 possible arrangements of the 5 digits. There are only two possible even digits, from which we will choose digit 1 and digit 5. The order of our choice certainly matters, since a different choice will give us a different 5-digit number. So we are choosing 2 digits from a set of 2 digits, where order matters. We therefore have P(2, 2) = 2*1 / 0! = 2 ways to choose these digits. The remaining 3 digits will comprise digits 2, 3 and 4. We are therefore choosing 3 digits from a set of 3, in order. There are P(3, 3) = 3*2*1/0! = 6 ways to do so. To obtain our number we can choose digits 1 and 5, then digits 2, 3 and 4. There are P(2, 2) * P(3, 3) = 2 * 6 = 12 ways to do this. So the probability that digits 1 and 5 are even is 12 / 120 = 1/10. Second way: A simpler solution looks at just the possibilities for digits 1 and 5. There are P(2, 2) = 2 choices for which these digits are even, and P(5, 2) = 20 total choices for these two digits. The probability that both will be even is therefore 2/20 = 1/10, the same as before. **
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RESPONSE --> ok
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QinwϏ̧I assignment #007 cʓ齯̀q} li Liberal Arts Mathematics II 10-11-2006
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19:35:36 12.2.6 single die, p(odd or <5). What is the probability of getting an odd result or a result < 5?
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RESPONSE --> odd number 3/6=0.5 <5 4/6=0.67
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19:37:12 ** there are 3 possible odd outcomes and four outcomes less that 5 which would add up to 7 outcomes, except that 2 of the outcomes < 5 are alrealdy odd and won't be counted. Thus the number of outcomes which are odd or less that 5 is 3 + 4 - 2 = 5 (this expresses the rule that n(A U B) = n(A) + n(B) - n(A ^ B), where U and ^ stand for union and intersection, respectively ). Thus the probability is 5/6. In terms of the specific sample space: The sample space for the experiment is {1, 2, 3, 4, 5, 6}. Success corresponds to events in the subset {1, 2, 3, 4, 5}. There are 6 elements in the sample space, 5 in the subset consisting of successful outcomes. Thus the probability is 5/6. **
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RESPONSE --> ok
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19:38:55 Query 12.2.15 drawing neither heart nor 7 from full deck
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RESPONSE --> probability- 9/13=0.69 odds- 9 to 4
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19:39:08 ** The sample space consists of the 52 cards in a full deck. There are 39 cards that aren't hearts, four 7's but only three aren't hearts so there are 36 cards that aren't hearts or seven. The probability is therefore 36/52 = 9/13. The odds in favor of the event are 16 to 36 (number favorable to number unfavorable), which in reduced form is 4 to 9. **
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RESPONSE --> ok
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19:42:41 12.2.24 prob of black flush or two pairs
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RESPONSE --> black flush=5108/2598960=0.001965 two pairs=123552/2598960=0.047539
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19:43:27 ** There are C(13,5) = 1287 ways to get a flush in a given suit--gotta choose the 5 cards from the 13 cards in that suit. There are two black suits so there are 2 * 1287 = 2574 possible black flushes. As the text tells you there are 123,552 ways to get two pairs. You can incidentally get this as 13 * C(4, 2) * 12 * C(4, 2) * C(44, 1) / 2 (2 of the 4 cards in any of the 13 denominations, then 2 of the 4 cards in any of the remaining 12 denominations, divide by 2 because the two denominations could occur in any order, then 1 of the 44 remaining cards not in either of the two denominations. There is no way that a hand can be both a black flush and two pairs, so there is no overlap to worry about (i.e., n(A and B) = 0 so n(A or B) = n(A) + n(B) - n(A and B) = n(A) + n(B) ). Thus there are 123,552 + 2574 = 126,126 ways to get one or the other. The probability is therefore 126,126 / 2,598,960 = .0485, approx. **
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RESPONSE --> ok
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19:45:30 12.2.33 x is sum of 2-digit numbers from {1, 2, ..., 5}; prob dist for random vbl x
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RESPONSE --> x P(x) 3 0.1 4 0.1 5 0.2 6 0.2 7 0.2 8 0.1 9 0.1
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19:45:41 ** If 2 different numbers are chosen from the set (1, 2, 3, 4, 5} then the sum 3 can appear only as 1+2. 4 can appear only as 1+3, assuming numbers can't be repeated (so, for example, 2+2 is not allowed). 5 can occur as 1+4 or as 2+3. 6 can occur as 1+5 or as 2+3. 7 can occur as 2+5 or as 1+6. 8 can occur only as 3+5. 9 can occur only as 4+5. Of the 10 possible combinations, the sums 3, 4, 8 and 9 can occur only once each, so each has probability .1. The sums 5, 6 and 7 can occur 2 times each, so each has probability .2. The possible sums are as indicated in the table below. 1 2 3 4 5 1 3 4 5 6 2 5 6 7 3 7 8 4 9 This assumes selection without replacement. There are C(5, 2) = 10 possible outcomes, as can be verified by counting the outcomes in the table. 3, 4, 8 and 9 appear once each as outcomes, so each has probability 1/10. 5, 6 and 7 appear twice each as outcomes, so each has probability 2/10. x p(x) 3 .1 4 .1 5 .2 6 .2 7 .2 8 .1 9 .1 **
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RESPONSE --> ok
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19:52:12 Query 12.2.36 n(A)=a, n(S) = s; P(A')=? What is the P(A')?
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RESPONSE --> It is the probability that an event will occur within sample space S
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19:52:32 ** A' is everything that is not in A. There are a ways A can happen, and s possibilities in the sample space S, so there are s - a ways A' can happen. So of the s possibilities, s-a are in A'. Thus the probability of A' is P(A') = n(A') / n(S) = (s - a) / s. **
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RESPONSE --> ok
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19:55:50 Query 12.2.42 spinners with 1-4 and 8-10; prob product is even
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RESPONSE --> 3/4=0.75
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19:57:05 ** The first number can be 1, 2, 3 or 4. The second can be 8, 9 or 10. There are therefore 4 * 3 = 12 possible outcomes. The only way to get an odd outcome is for the two numbers to both be odd. There are only 2 ways that can happen (1 * 9 and 3 * 9). The other 10 products are all even. So the probability of an even number is 10 / 12 = 5/6 = .833... . Alternatively we can set up the sample space in the form of the table 8 9 10 1 8 9 10 2 16 18 20 3 24 27 30 4 32 36 40 We see directly from this sample space that 10 of the 12 possible outcomes are even. **
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RESPONSE --> ok
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e}dQ~§ assignment #008 cʓ齯̀q} li Liberal Arts Mathematics II 10-11-2006
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20:08:47 Query 12.3.6 two members chosen for committee, Republican or no. Are the two choices independent or dependent and why?
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RESPONSE --> They could be dependent because if the first member chosen was republican, more than likely, the second will also be republican.
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20:08:57 ** The choice of the second is influenced by the first choice. If a Republican is chosen on the first choice, then there are fewer Republicans available for the second choice and the probability of getting a Republican on the second choice is lower than if a Republican had not been chosen first. COMMON ERROR: they are independant because they were randomly selected....if it is random then one did not depend on the other. EXPLANATION: The selection was indeed random, but the makeup of the remaining group available on the second choice depends on the first choice. **
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RESPONSE --> ok
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20:23:00 Query 12.3.12 table of motivations by male, female What is the probability that an individual will be primarily motivated by money or creativity given that the individual is female?
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RESPONSE --> 14/32=0.438 money 13/34=0.382 creativity
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20:23:32 ** There are 13 primarily motivated by money and 14 by creativity. Each person can have only one primary motivation so there is no overlap between these two groups. There are thus 13 + 14 = 27 motivated by money or creativity, out of a total of 66 women, which gives probability 27 / 66 = 9/22 = .41 approx.. If M is the set motivated by money, C the set motivated by creativity and S the entire sample space then we have p ( M or C) = p(M) + p(C) - p(M and C) or in set notation p(M U C) = p(M) + p(C ) - p(M ^ C), with U and ^ standing for set union and intersection. Since M ^ C is empty, p(M ^ C) = 0. p(M) = n(M) / n(S) = 13 / 66 and p(C) = n(C) / n(S) = 14/66 so we have p (M U C) = 13/66 + 14/66 = 27/66 = 9/22 = .41 approx. **
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RESPONSE --> ok, I figured the wrong things
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20:30:23 Query 12.3.24 prob of club 2d given diamond first What is the desired probability and how did you obtain it?
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RESPONSE --> There are four sets, club, diamond, spades, and hearts. If a diamond is dealt first, and club second, this probability is 1/4=0.25
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20:31:06 ** The probability is 13 / 51. When the second card is chosen there are 13 clubs still left, out of 51 remaining cards. This can also be calculated using P(A|B) = P(A^B) / P(B). The probability of getting a diamond on the first card and a club on the second is 13/52 * 13/51. The probability of getting a diamond on the first card is 13/52. So the probability of a club on the second given a diamond on the first is (13 / 52 * 13 / 51) / (13/52) = 13/51. **
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RESPONSE --> ok
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20:32:20 Query 12.3.32 prob of diamond given red What is the probability of getting a diamond given that the card is red?
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RESPONSE --> Well, it could either be a heart or a diamond, so it is 1/2, 0.50
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20:32:29 ** Of the 26 red cards, 13 are diamonds. So the probability of a diamond, given red, is 13/26 = 1/2. **
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RESPONSE --> ok
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20:38:34 Query 12.3.36 P(sale > $100) = .8; prob that first three sales all >$100 What is the probability that the first three sales are all for > $100 and how did you obtain your result?
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RESPONSE --> 0.3 you asked only for thee probability that all 3 sales would be >$100. That is 1/3=0.3
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20:38:49 ** the first sale has to be > $100, AND the second sale has to be > $100, AND the third sale has to be > $100. The events are independent. So the total probability, by the fundamental counting principle, is .8 * .8 * .8 = .512. **
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RESPONSE --> ok
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20:40:11 Query 12.3.42 P(critical direction) = .05; prob that none of the 5 scheduled days for launches has cloud movement in critical direction
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RESPONSE --> 0.015= 0.5^5
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20:40:25 ** On a given day cloud movement is not in critical direction with probability 1 - .05 = .95. this has to occur on the first day, then it has to occur on the second day, then on the third, then on the fourth, then on the fifth. These events are considered independent so the probability is .95 * .95 * .95 * .95 * .95 = .774 approx. (use your calculator to get the accurate answer). In order for none of the five days to have cloud movement in the critical direction, each of the five days must not have movement in the critical direction. The probability that the movement will not the in the critical direction for each of the days is .95. The probability that this will happen on every one of the five days is therefore .95 * .95 * .95 * .95 * .95 = .774, approx. **
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RESPONSE --> ok, I did it backward
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20:42:29 Query 12.3.54 probability of heads .52, tails .48; P(ht) What is the probability of head then tails?
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RESPONSE --> It is either 0.04 by finding the difference or 1
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20:42:44 ** There is a .52 probability of getting heads, then there is a .48 probability a getting tails. The two events have to happen consecutively. By the Fundamental Counting Principle there is thus a probability of .52 * .48 = .2496 of getting Heads then Tails. **
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RESPONSE --> ok
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20:45:17 Query 12.3.66 for given day P(rain)=.5, P(rain | rain day before) = .8, P(rain | no rain day before ) = .3. find P(rain on 3 consecutive days). For first 4 days in November what is the probability that it will rain on all four days given Oct 31 is clear?
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RESPONSE --> 0.04=0.5*0.8*0.3
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20:46:11 ** The probability of rain on the first of the four days is .3, since it is given that there was no rain on the previous day. The probability of rain on each of the following for days is .8, since on each of these days it rained the day before. The probability of rain on all four days is therefore .3 * .8 * .8 * .8 = .154. ANOTHER WAY OF SAYING IT: Oct 31 was clear so the probability of rain on the first day is .3. If it rained on the first day of the month then there is a probability of .8 that it rains on the second day. If it rained on the second day of the month then there is a probability of .8 that it rains on the third day. If it rained on the third day of the month then there is a probability of .8 that it rains on the fourth day. So the probability of rain on all 4 days is .3 * .8 * .8 * .8 = .154 **
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RESPONSE --> ok, I see how you got that. I needed to use the information given that 0.8 was the figure given for rain X 2 days
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20:47:19 Query 12.3.66 for given day P(rain)=.5, P(rain | rain day before) = .8, P(rain | no rain day before ) = .3. What is P(rain on 3 consecutive days).
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RESPONSE --> 0.8*0.3=0.24
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20:48:00 ** To get rain on 3 consecutive days requires rain on the first day, which happens with probability .5; then rain on the second day given that there was rain on the first day, which is .8, then rain on the third day, given that there was rain on the previous day; this third probability is also .8. The probability of the 3 events all happening (rain of 1st day AND rain on the second day AND rain on the third day) is therefore .5 * .8 * .8 = .32. **
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RESPONSE --> Ok, Why would you not use the day that rain had occurred but not the day before?
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20:48:09 QUESTION ON PROBLEM 33: Please explain Problem 33 of 12.3. It reads: If one number is chosen randomly from the intergers 1 throught 10, the probability of getting a number that is odd and prime, by the general multiplication rule is P(odd) * P(prime/odd) = 5/10 * 3/5 = 3/10 My question is how did we get three prime numbers out of 1 through 10? I assumed there were 4 of them (2, 3, 5, and 7).
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RESPONSE --> ok
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20:48:16 ** ONE ANSWER: The sample space is reduced to odd numbers, and 2 is not odd. So the set within the restricted sample space {1, 3, 5, 7, 9} should just be {3, 5, 7}. ANOTHER ANSWER: If we don't use the restricted sample space then we have P(prime | odd ) = P(prme) * P(odd | prime). We find P(prime) and P(odd | prime). P(prime) = 4 / 10, since there are 4 primes between 1 and 10. Within the unrestricted sample space P(odd | prime) is 3 / 4 since of the primes 2, 3, 5, 7 only three are odd. }Thus when you multiply P(prme) * P(odd | prime) you get 4/10 * 3/4 = 3/10, just as before. **
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RESPONSE --> ok
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20:48:44 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I didn't do as well with this section but I think with a little more practice, I should be fine.
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QP~GݴS{° assignment #009 cʓ齯̀q} li Liberal Arts Mathematics II 10-11-2006
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20:52:00 Query 12.4.3 P(2 H on 3 flips)
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RESPONSE --> 2/3*2/3*2/3=8/27=0.30
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20:52:31 ** On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of there 8 possibilities, only 3 of them have two Heads. Thus the probability is 3 / 8. You can get this result without listing. There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. If you have n flips, there are C(n,r) ways to get r Heads. This number appears in the n+1 row, as the r+1 entry, of Pascal's triangle. **
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RESPONSE --> ok
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20:53:52 What is the significance of .5^2 * .5 for this question?
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RESPONSE --> I'm not sure how this goes into the previous question
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20:54:54 ** .5^2 is the probability of getting Heads twice in a row. .5 is the probability of a Tails. .5^2 * .5 is therefore the probability of getting HHT. Since the probabilities are independent, you have the same probability of getting two Heads and a Tail in some different order. Since there are C(3,2) possible orders for 2 Heads on 3 coins, the probability of getting 2 Heads and one Tail is C(3,2) * .5^2 * .5 = 3 * .125 = .375, the same as the 3/8 we obtained by listing. **
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RESPONSE --> ok
.................... ִՉoDMj`
assignment #009 cʓ齯̀q} li Liberal Arts Mathematics II 11-27-2006
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23:59:08 Query 12.4.3 P(2 H on 3 flips)
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RESPONSE --> 3/8=0.375
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23:59:19 ** On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of there 8 possibilities, only 3 of them have two Heads. Thus the probability is 3 / 8. You can get this result without listing. There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. If you have n flips, there are C(n,r) ways to get r Heads. This number appears in the n+1 row, as the r+1 entry, of Pascal's triangle. **
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RESPONSE --> ok
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00:01:52 What is the significance of .5^2 * .5 for this question?
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RESPONSE --> it is the probability of getting heads twice in a row
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00:01:59 ** .5^2 is the probability of getting Heads twice in a row. .5 is the probability of a Tails. .5^2 * .5 is therefore the probability of getting HHT. Since the probabilities are independent, you have the same probability of getting two Heads and a Tail in some different order. Since there are C(3,2) possible orders for 2 Heads on 3 coins, the probability of getting 2 Heads and one Tail is C(3,2) * .5^2 * .5 = 3 * .125 = .375, the same as the 3/8 we obtained by listing. **
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RESPONSE --> ok
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00:03:43 Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> 7/8= 0.875 this is the probability of getting no heads on three flips
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00:03:51 ** Probability of getting no heads on three flips is P(TTT) = .5 * .5 * .5 = .125, or 1/8, obtained by multiplying the probability of getting a tails for each of 3 independent flips. Subtracting this from 1 gives .875, or 7/8. **
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RESPONSE --> ok
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00:06:22 Query 12.4.15 P(3 H on 7 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> c(7,3)=7*6*5/3= 35 ways 35/128
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00:06:28 ** There are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to choose three of the 7 `positions' for Heads on 7 flips. So there are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to get three heads on 7 flips. The probability of any of these ways is (1/2)^3 * (1/2)^4 = 1 / 2^7 = 1 / 128. The probability of 3 Heads on 7 flips is therefore 35 * 1/128 = 35 / 128. **
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RESPONSE --> ok
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00:08:39 Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die
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RESPONSE --> c(3,1)*1/6*(5,6)^2=4*1/6*25/36=100/216=25/72
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00:08:43 ** To get 1 success on 3 tries you have to get 1 success and 2 failures. On any flip the probability of success is 1/6 and the probability of failure is 5/6. For any ordered sequence with 1 success and 2 failures the probability is 1/6 * (5/6)^2. Since there are C(3,1) = 3 possible orders in which exactly 1 success can be obtained, the probability is C(3,1) * 1/6 * (5/6)^2 = 4 * 1/6 * 25 / 36 = 100 / 216 = 25 / 72. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/6, prob of failure q = 1 - 1/6 = 5/6, n = 3 trials and r = 1 success. **
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RESPONSE --> ok
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00:10:52 Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability?
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RESPONSE --> 1/3 (1/3)^7*(2/3)^3 c(10,7) 320/19683=0.016
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00:10:56 ** The probability of a correct answer from a random choice on any single question is 1/3. For any sequence of 7 correct answers and 3 incorrect the probability is (1/3)^7 * (2/3)^3. There are C(10,7) possible positions for 7 correct answers among 10 questions. So the probability is C(10,7) * (1/3)^7 * (2/3)^3 = 320/19683 = 0.0163 approx. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/3, prob of failure q = 1 - 1/3 = 2/3, n = 10 trials and r = 7 success. ANOTHER SOLUTION: There are C(10,7) ways to distribute the 7 correct answers among the 10 questions. The probability of any single outcome with 7 successes and 3 failures is the product of (1/3)^7, representing 7 successes, and (2/3)^3, representing 3 failures. The probability of exactly seven correct questions is therefore prob = C(10,7) * (2/3)^3 * (1/3)^7 . **
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RESPONSE --> ok
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00:12:11 Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each
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RESPONSE --> 0.448
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00:12:18 ** The probability of 0 side effects is C(8,0) * .7^8. The probability of 1 side effect is C(8,1) * .7^7 * .3^1. The probability of 2 side effects is C(8,2) * .7^6 * .3^2. The sum of these two probabilities is the probability that two or fewer patients will have side effects. We subtract this probability from 1 to get the probability that more than 2 will experience side effects. The result is approximately .448. DER**
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RESPONSE --> ok
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00:13:10 Query 12.4.48 P(4 th child is 1 st daughter) What is the probability that the fourth child is the first daughter and how did you obtain your result?
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RESPONSE --> 1/2*1/2*1/2*1/2=0.0625
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00:13:16 ** The fourth child will be the first daughter if the sequence is SSSD, S standing for son and D for daughter. The probability of S on any birth is .5, and the probability of G is .5. The probability of SSSD is .5^3 * .5 = .0625 or 1/16. **
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RESPONSE --> ok
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00:15:50 Query 12.4.54 10-step rnd walk, 1 dim; P(6 South) What is the probability of ending up 6 blocks South of the starting point and how did you obtain it?
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RESPONSE --> 1/10*1/6=0.016
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00:15:57 ** To end up 6 blocks South requires 8 steps South and 2 steps North. Thus exactly 8 of the 10 steps must be South, and there are C(10,8) ways for this to happen. The probability of any given combination of 8 South and 2 North is (1/2)^8 * (1/2)^2 = 1 / 2^10 = 1 / 1024. The probability of ending up 6 blocks South is therefore prob = C(10,8) * (1/2)^8 * (1/2)^2 = 45 * (1/2)^10 = 45 / 1024, or about .044. **
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RESPONSE --> ok
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00:16:09 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
........................................................................................................................................................... خ܀Jé{җ˨[X͂
assignment #010 cʓ齯̀q} li Liberal Arts Mathematics II 11-28-2006
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00:24:16 Query 12.5.6 fair dice game pays $3 for 6, $2 for 5, $1 for 4. What is a fair price to pay for playing this game?
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RESPONSE --> 3.00 per game
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00:24:30 ** A 1 in 6 chance of getting $3 is worth 1/6 * $3 = $.50 . A 1 in 6 chance of getting $2 is worth 1/6 * $2 = $.33 1/3 . A 1 in 6 chance of getting $1 is worth 1/6 * $1 = $.16 2/3 . The total expectation is $1.00 * 1/6 + $2.00 * 1/6 + $3.00 * 1/6 = $1.00 So a fair price to pay is $1.00 **
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RESPONSE --> ok
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00:28:00 Query 12.5.10 expectation Roulette $1 bet 18 red, 18 black one zero What is the expected net value of a bet on red?
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RESPONSE --> ($1)18/37=0.0486
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00:28:13 ** If your net gain is $1 for a win and -$1 for a loss the expected value is 18/37 * (+1) + 19/37 * (-1) = -$.027. **
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RESPONSE --> ok
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00:30:07 Query 12.5.20 exp sum of 2 of 5 cards 1-5. What is the expected sum of the numbers on the two cards drawn?
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RESPONSE --> 1/5+1/5=2/5
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00:30:17 ** You can't get a sum of 1 on two cards. There is also no way to get a sum of two, since the lowest total possible is 1 + 2 = 3. There are 2 ways to get total 3. You can get 1 on the first and 2 on the second, or vice versa. There are 2 ways to get total 4. You can get 1 on the first and 3 on the second, or vice versa. There are 4 ways to get total 5. You can get 1 on the first and 4 on the second, or vice versa, or 2 on the first and 3 on the second, or vice versa. There are 4 ways to get total 6. You can get 1 on the first and 5 on the second, or vice versa, or 2 on the first and 4 on the second, or vice versa. There are 4 ways to get total 7. You can get 2 on the first and 5 on the second, or vice versa, or 4 on the first and 3 on the second, or vice versa. There are 2 ways to get total 8. You can get 3 on the first and 5 on the second, or vice versa. There are 2 ways to get total 9. You can get 4 on the first and 5 on the second, or vice versa. You can't get more than 9. There are 2+2+4+4+4+2+2 = 20 possibilities, so the probabilities are 2/20, 4/20, 5/20, etc.. The expected sum is therefore 2/20 * 3 + 2/20 * 4 + 4/20 * 5 + 4/20 * 6 + 4/20 * 7 + 2/20 * 8 + 2/20 * 9. This gives 120 / 20 = 6. **
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RESPONSE --> ok
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00:30:23 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ok
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Ƶw׆wǪ assignment #011 cʓ齯̀q} li Liberal Arts Mathematics II 11-28-2006
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00:38:55 **** Query 12.6.6 rnd # table to simulate 50 one-and-one foul shooting opportunities if 70% prob of success; 2 shots Give the results of your tally. How does your empirical probability compare with the theoretical probability?
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RESPONSE --> Pts. Tally 0 2 1 0 2 0
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00:39:10 ** In 1-and-1 shooting you only get a second shot if you make the first. So there are 3 possibilities: miss the first, don't get another shot make the first, get another shot and make it make the first, get another shot and miss it prob of 0 = prob of miss on first shot = .3 prob of 1 = prob of hit on first and miss on 2d = .3 * .7 = .21 prob of 2 = prob ot hit * prob of hit = .49. 'Hits' happen with 70% or .7 probability, misses with probability 30% or 3. The theoretical probability of 2 misses is probability of miss * probability of miss = .30 * .30 = .09. The theoretical probability of 2 miss and 1 hit is probability of miss * probability of hit + probability of miss *hit probability of miss = .30 * .70 + .70 * .30 = .21 + .21 = .42. The theoretical probability of 2 hits is probability of hit * probability of hit = .70 * .70 = .49. Note that these probabilities add up to .09 + .42 + .49 = 1, as they must since these three events cover all possibilities. To use the table, randomly pick a starting point. Let numbers 1-7 correspond to making the free throw, with 8, 9 and 0 corresponding to misses. Go down the list, or across the list in an order you decided before looking at the list. Read two digits from the list and see if they correspond to two 'hits', two 'misses' or a 'hit' and a 'miss'. Record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Read two more digits and record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Continue until you have the required number of results. Tally how many times you got 0 'hits', 1 'hit', 2 'hits' etc.. Any outcome that starts with a 'miss' corresponds to zero point. 'Hit-miss' corresopnds to 1 point and 'hit-hit' corresponds to 2 points. Determine the percent of time you got each number of points, and compare to the theoretical probabilities .09, .42 and .49. *&*& **
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RESPONSE --> ok
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00:43:40 Query 12.6.12 rnd walk start N then right, left or straight with prob 1/2, 1/6, 1/3; 1 st 2 columns of table
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RESPONSE --> 1/2*1/6*1/3=0.02
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00:45:12 ** Your probabilities are given as 1/2, 1/6 and 1/3. These can all be expressed in terms of the common denominator 6: 1/2 = 3/6, 1/6 = 1/6, 1/3 = 2/6. So a move to the right has 3 chances out of 6, a move to the left has 1 chance out of 6 and a move straight has 2 chances out of 6. You can simulate this by letting the three digits 1, 2, 3 stand for a move to the right, the single digit 4 for a move to the left and the two digits 5, 6 for a straight move. The remaining digits 0, 7, 8, 9 don't stand for anything, and if you land on one of these numbers you just move to the next number. So according to your the first two columns of you table, how many times do you move to the right, how many to the left, how many straight and where do you end up? **
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RESPONSE --> 4 lefts, 4 straights back at the beginning?
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00:45:24 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I'm completely confused
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00:45:34 014. `query 14
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RESPONSE --> ok
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00:52:11 query problem 13.1.6 freq dist 35 IQ scores class width 5
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RESPONSE --> I'm not sure if I understand this but if I did it correctly on paper, the graph shows an inrease in IQ scores, starting at 91 and going all the way to 134
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00:52:53 ** These numbers might or might not be completely accurate, but the calculation of the relative frequencies as percents follows from the tallies: 91-95 1 1/50 = 2% 96-100 3 3/50 = 6% 101-105 5 5/50 = 10% 106-110 7 7/50 = 14% 111-115 12 12/50 = 24% 116-120 9 9/50 = 18% 121-125 8 8/50 = 16% 126-130 3 3/50 = 6% 131-135 2 2/50 = 4% **
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RESPONSE --> oh, I see how it goes now
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01:04:26 query problem 13.1.10 stem and leaf for yards gained by 44 rushers
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RESPONSE --> 0 3,7 1 2,6,8,9 2 2,4,5,8,9 3 0,2,3,3,6,6,7,9 4 0,1,2,3,3,5,6,9 5 1,4,5,8 6 0,2,7 7 3,3,9 8 6,8 9 4 10 2 11 2 12 3
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01:04:40 ** In order we get the following: 0 3 7 1 2 6 8 9 2 2 4 5 8 9 9 3 0 2 3 3 6 6 7 9 4 0 1 2 3 3 5 6 9 5 1 4 5 8 6 0 2 7 7 3 3 9 8 6 8 9 4 10 2 11 2 12 3 **
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RESPONSE --> ok
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01:06:26 query problem 13.1.35 empirical probability distribution for letters of alphabet
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RESPONSE --> letter probability A 0.208 E 0.338 I 0.169 O 0.208 U 0.078
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01:06:55 ** The probabilities for A, E, I, O, U are: .08. .13, .065, .08, .03. The sum of these probabilities is .385 The probabilities of the letters given a vowel: .08 / .385 = .208 .13 / .385 = .338 .065 / .385 = .169 .08 / .385 = .208 .03 / .385 = .078 These probabilities total 1. **
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RESPONSE --> OK
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01:07:05 015. `query 15
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RESPONSE --> OK
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01:11:29 query problem 13.2.10 .3, .4, .3, .8, .7, .9, .2, .1, .5, .9, .6
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RESPONSE --> MEAN- 0.52 MEDIAN- 0.5 MODE- 0.3, 0.9
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01:11:54 ** The numbers, in order, are .1, .2, .3, .3, .4, .5, .6, .7, .8, .9, .9 The mean, obtained by adding the 11 numbers then dividing by 11, is .518. The median occurs at position (n + 1 ) / 2 = 6 in the ordered list. This number is .5. Note that there are five numbers before .5 and five numbers after .5. The maximum number of times a number repeats in this distribution is 2. So there are two modes (and we say that the distribution is bimodal). The modes are .3 and .9. **
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RESPONSE --> I miscounted
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01:16:50 **** query problem 13.2.24 more effect from extreme value
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RESPONSE --> the mean would show more of an effect
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01:17:01 ** The mean is drastically affected by the error; correcting the error changes the mean by about 3 units. The median number, however, simply shifts 1 position, changing from 2.28 to 2.39. **
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RESPONSE --> ok
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01:21:43 **** query problem 13.2.30 Salaries 6 @$19k, 8 @ 23k, 2 @ 34.5k, 7 @ 56.9k, 1 @ 145.5k.
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RESPONSE --> the mean is 135,000
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01:22:03 ** IF THERE ARE 28 EMPLOYEES: The totals paid for each salary level are: 6 * $19,500 = $117,000 8 * $23,000 = $184,000 4 * $28,300 = $113,200 2 * $34,500 = $69,000 7 * $36,900 = $258,300 1 * $145,500 = $145,500 The grand total paid in salaries to the 28 employees is therefore $887,000, giving an average of $887,000 / 28 = $31,700. The median occurs at position (n + 1) / 2 = (28 + 1) / 2 = 14.5. Since the 14 th salaray on a list ordered from least to greatest is $23,000 and the 15 th is $28300 the median is ($23000 +$28300) / 2 = $25,650. The mode is 23,000, since this salary occurs more frequently than any other. IF THERE ARE 24 EMPLOYEES: The totals paid for each salary level are: $19,000 * 6 = $114,000 $23,000 * 8 = $184,000 $34,500 * 2 = $69,000 $56,900 * 7 = $398,300 $145,500 * 1 = $145,500 Adding these gives a grand total, which is divided by the number 24 of employees to obtain the mean $37,950. The median occurs at position (n + 1) / 2 = (24 + 1) / 2 = 12.5. Since the $23000 salary covers positions 7 thru 14 in an ordered lise of salaries the median is $23,000. The mode is 23,000, since this salary occurs more frequently than any other.
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RESPONSE --> ok
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01:22:56 **** query problem 13.2.51 mean, med, mode of 0, 1, 3, 14, 14, 15, 16, 16, 17, 17, 18, 18, 18, 19, 20
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RESPONSE --> mean- 13.7 median- 16 mode- 18
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01:23:02 ** The mean is 13.73, obtained by adding together all the numbers and dividing by n = 15. The median is in position (n+1) / 2 = (15+1)/2 = 8 on the ordered list; the 8 th number is 16. The mode is 18, which is the only number occurring as many as 3 times. **
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RESPONSE --> ok
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01:23:07 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ok
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01:23:24 016. `query 16
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RESPONSE --> ok
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01:31:18 query probl 13.3.6 range, std dev of {67, 83, 55, 68, 77, 63, 84, 72, 65}
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RESPONSE --> range-84-55=29 mean- 70.4 standard deviation- 55-70.4= -15.4 63-70.4= -7.4 65-70.4= -5.4 67-70.4= -3.4 68-70.4= -2.4 72-70.4= 1.6 77-70.4= 6.6 83-70.4= 12.6 84-70.4= 13.6
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01:31:42 ** x dev. from mean squared dev. 55 15.4 237.16 63 7.4 54.76 65 5.4 29.16 67 3.4 11.56 68 2.4 5.76 72 1.6 2.56 77 6.6 43.56 83 12.6 158.76 84 13.6 184.96 634 728.08 mean = 634 / 9 = 70.4 std. dev. = `sqrt (728.08 / 8) = 9.54 range = 84 - 55 = 29 **
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RESPONSE --> ok
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01:49:19 **** query probl 13.3.12 freq dist 14,8; 16,12; 18,15; 20,14; 22,10; 24,6; 26,3
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RESPONSE --> 6 4 3 6 12 18 23 mean-10.3 6-10.3=-4.3 18.49 4-10.3=-6.3 39.69 3-10.3=-7.3 53.29 6-10.3=-4.3 18.49 12-10.3=1.7 2.89 18-10.3=7.7 59.29 23-10.3=12.7 161.29
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01:49:33 ** Value freq Value * Freq Dev^2 * F 14 8 112 204.80 16 12 192 112.32 18 15 270 16.80 20 14 280 12.32 22 10 220 86.40 24 6 144 146.40 26 3 78 144.48 Total 68 1296 723.52 Total squared dev is 723.5 so ave squared dev is 723.5 / 68 = 10.6, approx. Std dev is sqrt(ave squared dev) = sqrt(10.6) = 3.3 approx. **
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RESPONSE --> ok
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01:54:14 **** query probl 13.3.18 chebyshev for z=5 What is the least possible number of elements of a sample which lie within 5 standard deviations of the mean?
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RESPONSE --> 1- 2/5= 0.60
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01:54:28 ** The formula 1 - 1/k^2 gave you .96. That's the proportion which must under any circumstances lie between mean and 5 std dev from the mean. So the number is .96 n, where n is the number of elements in the sample. **
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RESPONSE --> ok
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01:55:06 query probl 13.3.48 mean length of stay 2.7 days, std dev 7.1 days.
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RESPONSE -->
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01:55:12 ** A sketch of a normal distribution will be a normal, or bell-shaped curve with its peak at the mean, dropping to about 60% of peak value at 1 std dev from the mean and to about 14% of peak value at 2 std dev from the mean. For a normal distribution with mean 2.7 and std dev. 7.1 one std dev from the mean occurs at 2.7 + 7.1 = 9.8 and at 2.7 7.1 = -4.4; two std dev from the mean occurs at 16.9 and -11.5. The corresponding normal curve cannot represent length of stay, since length of stay must not be less than zero. As a result we obtain a curve which tails off for large values of x, but whose area is concentrated mostly between 0 and 2.7. This curve is not symmetric like the bell curve but is very skewed, bunched up on one side of the mean 2.7 and more spread out for larger values. **
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RESPONSE --> ok
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01:56:05 **** Describe your sketch of the distribution of lengths of stay.
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RESPONSE --> The bell shape would go from 2.7 days at the lowest and 7.1 days at the highest
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01:56:18 2.7 is in the center with each number within 7.1 of the right or left of 2.7 and each additonal number on the left or right within 7.1 of each other. I see the curve as not being skewed.
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RESPONSE --> ok
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01:57:10 **** Is your distribution skewed? If so why, and if not why do you think it shouldn't be?
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RESPONSE --> yes, because of the deviation of 7.1
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01:57:18 STUDENT RESPONSE AND INSTRUCTOR COMMENT: I said no, but I'm not really sure how to determine this if I draw it myself, becasue I place all the numbers at an equal distance from each other. It just hit me while typing this that it must be skewed to the right because I can't really have negative days stay in the hospital. I think I have confused myself ** You didn't confuse yourself. That's exactly the point. You can't stay fewer than 0 days; since even 1 standard deviation is way below 0 the deviations must be primarily to the right of the mean. So the distribution must be skewed significantly to the right. GENERAL SUGGESTION: In general to understand the graphs of various distributions, try to understand in terms first of the bell-shaped curve with max height at the mean, dropping to about 60% height at a distance of 1 std dev from the mean and to about 14% at 2 std dev from the mean. Then understand that this distribution can be distorted, or skewed, as in this problem. This occurs when most of the distribution lies close to the mean on one side, with a smaller part of the distribution spread out further from the mean on the other. **
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RESPONSE --> ok
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01:57:38 017. `query 17
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RESPONSE --> ok
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course Mth 152 Here is the rest of the assignments. I went to take the tests and the lab was closed. I will try to take them tomorrow and if I can't, i will have them done on Monday. ִՉoDMj`assignment #009
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23:59:08 Query 12.4.3 P(2 H on 3 flips)
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RESPONSE --> 3/8=0.375
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23:59:19 ** On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of there 8 possibilities, only 3 of them have two Heads. Thus the probability is 3 / 8. You can get this result without listing. There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. If you have n flips, there are C(n,r) ways to get r Heads. This number appears in the n+1 row, as the r+1 entry, of Pascal's triangle. **
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RESPONSE --> ok
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00:01:52 What is the significance of .5^2 * .5 for this question?
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RESPONSE --> it is the probability of getting heads twice in a row
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00:01:59 ** .5^2 is the probability of getting Heads twice in a row. .5 is the probability of a Tails. .5^2 * .5 is therefore the probability of getting HHT. Since the probabilities are independent, you have the same probability of getting two Heads and a Tail in some different order. Since there are C(3,2) possible orders for 2 Heads on 3 coins, the probability of getting 2 Heads and one Tail is C(3,2) * .5^2 * .5 = 3 * .125 = .375, the same as the 3/8 we obtained by listing. **
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RESPONSE --> ok
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00:03:43 Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> 7/8= 0.875 this is the probability of getting no heads on three flips
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00:03:51 ** Probability of getting no heads on three flips is P(TTT) = .5 * .5 * .5 = .125, or 1/8, obtained by multiplying the probability of getting a tails for each of 3 independent flips. Subtracting this from 1 gives .875, or 7/8. **
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RESPONSE --> ok
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00:06:22 Query 12.4.15 P(3 H on 7 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> c(7,3)=7*6*5/3= 35 ways 35/128
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00:06:28 ** There are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to choose three of the 7 `positions' for Heads on 7 flips. So there are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to get three heads on 7 flips. The probability of any of these ways is (1/2)^3 * (1/2)^4 = 1 / 2^7 = 1 / 128. The probability of 3 Heads on 7 flips is therefore 35 * 1/128 = 35 / 128. **
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RESPONSE --> ok
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00:08:39 Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die
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RESPONSE --> c(3,1)*1/6*(5,6)^2=4*1/6*25/36=100/216=25/72
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00:08:43 ** To get 1 success on 3 tries you have to get 1 success and 2 failures. On any flip the probability of success is 1/6 and the probability of failure is 5/6. For any ordered sequence with 1 success and 2 failures the probability is 1/6 * (5/6)^2. Since there are C(3,1) = 3 possible orders in which exactly 1 success can be obtained, the probability is C(3,1) * 1/6 * (5/6)^2 = 4 * 1/6 * 25 / 36 = 100 / 216 = 25 / 72. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/6, prob of failure q = 1 - 1/6 = 5/6, n = 3 trials and r = 1 success. **
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RESPONSE --> ok
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00:10:52 Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability?
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RESPONSE --> 1/3 (1/3)^7*(2/3)^3 c(10,7) 320/19683=0.016
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00:10:56 ** The probability of a correct answer from a random choice on any single question is 1/3. For any sequence of 7 correct answers and 3 incorrect the probability is (1/3)^7 * (2/3)^3. There are C(10,7) possible positions for 7 correct answers among 10 questions. So the probability is C(10,7) * (1/3)^7 * (2/3)^3 = 320/19683 = 0.0163 approx. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/3, prob of failure q = 1 - 1/3 = 2/3, n = 10 trials and r = 7 success. ANOTHER SOLUTION: There are C(10,7) ways to distribute the 7 correct answers among the 10 questions. The probability of any single outcome with 7 successes and 3 failures is the product of (1/3)^7, representing 7 successes, and (2/3)^3, representing 3 failures. The probability of exactly seven correct questions is therefore prob = C(10,7) * (2/3)^3 * (1/3)^7 . **
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RESPONSE --> ok
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00:12:11 Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each
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RESPONSE --> 0.448
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00:12:18 ** The probability of 0 side effects is C(8,0) * .7^8. The probability of 1 side effect is C(8,1) * .7^7 * .3^1. The probability of 2 side effects is C(8,2) * .7^6 * .3^2. The sum of these two probabilities is the probability that two or fewer patients will have side effects. We subtract this probability from 1 to get the probability that more than 2 will experience side effects. The result is approximately .448. DER**
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RESPONSE --> ok
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00:13:10 Query 12.4.48 P(4 th child is 1 st daughter) What is the probability that the fourth child is the first daughter and how did you obtain your result?
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RESPONSE --> 1/2*1/2*1/2*1/2=0.0625
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00:13:16 ** The fourth child will be the first daughter if the sequence is SSSD, S standing for son and D for daughter. The probability of S on any birth is .5, and the probability of G is .5. The probability of SSSD is .5^3 * .5 = .0625 or 1/16. **
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RESPONSE --> ok
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00:15:50 Query 12.4.54 10-step rnd walk, 1 dim; P(6 South) What is the probability of ending up 6 blocks South of the starting point and how did you obtain it?
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RESPONSE --> 1/10*1/6=0.016
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00:15:57 ** To end up 6 blocks South requires 8 steps South and 2 steps North. Thus exactly 8 of the 10 steps must be South, and there are C(10,8) ways for this to happen. The probability of any given combination of 8 South and 2 North is (1/2)^8 * (1/2)^2 = 1 / 2^10 = 1 / 1024. خ܀Jé{җ˨[X͂ assignment #010 cʓ齯̀q} li Liberal Arts Mathematics II 11-28-2006
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00:24:16 Query 12.5.6 fair dice game pays $3 for 6, $2 for 5, $1 for 4. What is a fair price to pay for playing this game?
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RESPONSE --> 3.00 per game
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00:24:30 ** A 1 in 6 chance of getting $3 is worth 1/6 * $3 = $.50 . A 1 in 6 chance of getting $2 is worth 1/6 * $2 = $.33 1/3 . A 1 in 6 chance of getting $1 is worth 1/6 * $1 = $.16 2/3 . The total expectation is $1.00 * 1/6 + $2.00 * 1/6 + $3.00 * 1/6 = $1.00 So a fair price to pay is $1.00 **
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RESPONSE --> ok
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00:28:00 Query 12.5.10 expectation Roulette $1 bet 18 red, 18 black one zero What is the expected net value of a bet on red?
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RESPONSE --> ($1)18/37=0.0486
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00:28:13 ** If your net gain is $1 for a win and -$1 for a loss the expected value is 18/37 * (+1) + 19/37 * (-1) = -$.027. **
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RESPONSE --> ok
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00:30:07 Query 12.5.20 exp sum of 2 of 5 cards 1-5. What is the expected sum of the numbers on the two cards drawn?
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RESPONSE --> 1/5+1/5=2/5
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00:30:17 ** You can't get a sum of 1 on two cards. There is also no way to get a sum of two, since the lowest total possible is 1 + 2 = 3. There are 2 ways to get total 3. You can get 1 on the first and 2 on the second, or vice versa. There are 2 ways to get total 4. You can get 1 on the first and 3 on the second, or vice versa. There are 4 ways to get total 5. You can get 1 on the first and 4 on the second, or vice versa, or 2 on the first and 3 on the second, or vice versa. There are 4 ways to get total 6. You can get 1 on the first and 5 on the second, or vice versa, or 2 on the first and 4 on the second, or vice versa. There are 4 ways to get total 7. You can get 2 on the first and 5 on the second, or vice versa, or 4 on the first and 3 on the second, or vice versa. There are 2 ways to get total 8. You can get 3 on the first and 5 on the second, or vice versa. There are 2 ways to get total 9. You can get 4 on the first and 5 on the second, or vice versa. You can't get more than 9. There are 2+2+4+4+4+2+2 = 20 possibilities, so the probabilities are 2/20, 4/20, 5/20, etc.. The expected sum is therefore 2/20 * 3 + 2/20 * 4 + 4/20 * 5 + 4/20 * 6 + 4/20 * 7 + 2/20 * 8 + 2/20 * 9. This gives 120 / 20 = 6. **
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RESPONSE --> ok
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00:30:23 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ok
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Ƶw׆wǪ assignment #011 cʓ齯̀q} li Liberal Arts Mathematics II 11-28-2006
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00:38:55 **** Query 12.6.6 rnd # table to simulate 50 one-and-one foul shooting opportunities if 70% prob of success; 2 shots Give the results of your tally. How does your empirical probability compare with the theoretical probability?
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RESPONSE --> Pts. Tally 0 2 1 0 2 0
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00:39:10 ** In 1-and-1 shooting you only get a second shot if you make the first. So there are 3 possibilities: miss the first, don't get another shot make the first, get another shot and make it make the first, get another shot and miss it prob of 0 = prob of miss on first shot = .3 prob of 1 = prob of hit on first and miss on 2d = .3 * .7 = .21 prob of 2 = prob ot hit * prob of hit = .49. 'Hits' happen with 70% or .7 probability, misses with probability 30% or 3. The theoretical probability of 2 misses is probability of miss * probability of miss = .30 * .30 = .09. The theoretical probability of 2 miss and 1 hit is probability of miss * probability of hit + probability of miss *hit probability of miss = .30 * .70 + .70 * .30 = .21 + .21 = .42. The theoretical probability of 2 hits is probability of hit * probability of hit = .70 * .70 = .49. Note that these probabilities add up to .09 + .42 + .49 = 1, as they must since these three events cover all possibilities. To use the table, randomly pick a starting point. Let numbers 1-7 correspond to making the free throw, with 8, 9 and 0 corresponding to misses. Go down the list, or across the list in an order you decided before looking at the list. Read two digits from the list and see if they correspond to two 'hits', two 'misses' or a 'hit' and a 'miss'. Record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Read two more digits and record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Continue until you have the required number of results. Tally how many times you got 0 'hits', 1 'hit', 2 'hits' etc.. Any outcome that starts with a 'miss' corresponds to zero point. 'Hit-miss' corresopnds to 1 point and 'hit-hit' corresponds to 2 points. Determine the percent of time you got each number of points, and compare to the theoretical probabilities .09, .42 and .49. *&*& **
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RESPONSE --> ok
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00:43:40 Query 12.6.12 rnd walk start N then right, left or straight with prob 1/2, 1/6, 1/3; 1 st 2 columns of table
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RESPONSE --> 1/2*1/6*1/3=0.02
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00:45:12 ** Your probabilities are given as 1/2, 1/6 and 1/3. These can all be expressed in terms of the common denominator 6: 1/2 = 3/6, 1/6 = 1/6, 1/3 = 2/6. So a move to the right has 3 chances out of 6, a move to the left has 1 chance out of 6 and a move straight has 2 chances out of 6. You can simulate this by letting the three digits 1, 2, 3 stand for a move to the right, the single digit 4 for a move to the left and the two digits 5, 6 for a straight move. The remaining digits 0, 7, 8, 9 don't stand for anything, and if you land on one of these numbers you just move to the next number. So according to your the first two columns of you table, how many times do you move to the right, how many to the left, how many straight and where do you end up? **
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RESPONSE --> 4 lefts, 4 straights back at the beginning?
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00:45:24 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I'm completely confused
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00:45:34 014. `query 14
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RESPONSE --> ok
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00:52:11 query problem 13.1.6 freq dist 35 IQ scores class width 5
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RESPONSE --> I'm not sure if I understand this but if I did it correctly on paper, the graph shows an inrease in IQ scores, starting at 91 and going all the way to 134
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00:52:53 ** These numbers might or might not be completely accurate, but the calculation of the relative frequencies as percents follows from the tallies: 91-95 1 1/50 = 2% 96-100 3 3/50 = 6% 101-105 5 5/50 = 10% 106-110 7 7/50 = 14% 111-115 12 12/50 = 24% 116-120 9 9/50 = 18% 121-125 8 8/50 = 16% 126-130 3 3/50 = 6% 131-135 2 2/50 = 4% **
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RESPONSE --> oh, I see how it goes now
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01:04:26 query problem 13.1.10 stem and leaf for yards gained by 44 rushers
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RESPONSE --> 0 3,7 1 2,6,8,9 2 2,4,5,8,9 3 0,2,3,3,6,6,7,9 4 0,1,2,3,3,5,6,9 5 1,4,5,8 6 0,2,7 7 3,3,9 8 6,8 9 4 10 2 11 2 12 3
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01:04:40 ** In order we get the following: 0 3 7 1 2 6 8 9 2 2 4 5 8 9 9 3 0 2 3 3 6 6 7 9 4 0 1 2 3 3 5 6 9 5 1 4 5 8 6 0 2 7 7 3 3 9 8 6 8 9 4 10 2 11 2 12 3 **
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RESPONSE --> ok
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01:06:26 query problem 13.1.35 empirical probability distribution for letters of alphabet
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RESPONSE --> letter probability A 0.208 E 0.338 I 0.169 O 0.208 U 0.078
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01:06:55 ** The probabilities for A, E, I, O, U are: .08. .13, .065, .08, .03. The sum of these probabilities is .385 The probabilities of the letters given a vowel: .08 / .385 = .208 .13 / .385 = .338 .065 / .385 = .169 .08 / .385 = .208 .03 / .385 = .078 These probabilities total 1. **
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RESPONSE --> OK
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01:07:05 015. `query 15
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RESPONSE --> OK
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01:11:29 query problem 13.2.10 .3, .4, .3, .8, .7, .9, .2, .1, .5, .9, .6
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RESPONSE --> MEAN- 0.52 MEDIAN- 0.5 MODE- 0.3, 0.9
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01:11:54 ** The numbers, in order, are .1, .2, .3, .3, .4, .5, .6, .7, .8, .9, .9 The mean, obtained by adding the 11 numbers then dividing by 11, is .518. The median occurs at position (n + 1 ) / 2 = 6 in the ordered list. This number is .5. Note that there are five numbers before .5 and five numbers after .5. The maximum number of times a number repeats in this distribution is 2. So there are two modes (and we say that the distribution is bimodal). The modes are .3 and .9. **
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RESPONSE --> I miscounted
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01:16:50 **** query problem 13.2.24 more effect from extreme value
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RESPONSE --> the mean would show more of an effect
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01:17:01 ** The mean is drastically affected by the error; correcting the error changes the mean by about 3 units. The median number, however, simply shifts 1 position, changing from 2.28 to 2.39. **
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RESPONSE --> ok
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01:21:43 **** query problem 13.2.30 Salaries 6 @$19k, 8 @ 23k, 2 @ 34.5k, 7 @ 56.9k, 1 @ 145.5k.
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RESPONSE --> the mean is 135,000
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01:22:03 ** IF THERE ARE 28 EMPLOYEES: The totals paid for each salary level are: 6 * $19,500 = $117,000 8 * $23,000 = $184,000 4 * $28,300 = $113,200 2 * $34,500 = $69,000 7 * $36,900 = $258,300 1 * $145,500 = $145,500 The grand total paid in salaries to the 28 employees is therefore $887,000, giving an average of $887,000 / 28 = $31,700. The median occurs at position (n + 1) / 2 = (28 + 1) / 2 = 14.5. Since the 14 th salaray on a list ordered from least to greatest is $23,000 and the 15 th is $28300 the median is ($23000 +$28300) / 2 = $25,650. The mode is 23,000, since this salary occurs more frequently than any other. IF THERE ARE 24 EMPLOYEES: The totals paid for each salary level are: $19,000 * 6 = $114,000 $23,000 * 8 = $184,000 $34,500 * 2 = $69,000 $56,900 * 7 = $398,300 $145,500 * 1 = $145,500 Adding these gives a grand total, which is divided by the number 24 of employees to obtain the mean $37,950. The median occurs at position (n + 1) / 2 = (24 + 1) / 2 = 12.5. Since the $23000 salary covers positions 7 thru 14 in an ordered lise of salaries the median is $23,000. The mode is 23,000, since this salary occurs more frequently than any other.
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RESPONSE --> ok
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01:22:56 **** query problem 13.2.51 mean, med, mode of 0, 1, 3, 14, 14, 15, 16, 16, 17, 17, 18, 18, 18, 19, 20
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RESPONSE --> mean- 13.7 median- 16 mode- 18
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01:23:02 ** The mean is 13.73, obtained by adding together all the numbers and dividing by n = 15. The median is in position (n+1) / 2 = (15+1)/2 = 8 on the ordered list; the 8 th number is 16. The mode is 18, which is the only number occurring as many as 3 times. **
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RESPONSE --> ok
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01:23:07 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ok
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01:23:24 016. `query 16
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RESPONSE --> ok
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01:31:18 query probl 13.3.6 range, std dev of {67, 83, 55, 68, 77, 63, 84, 72, 65}
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RESPONSE --> range-84-55=29 mean- 70.4 standard deviation- 55-70.4= -15.4 63-70.4= -7.4 65-70.4= -5.4 67-70.4= -3.4 68-70.4= -2.4 72-70.4= 1.6 77-70.4= 6.6 83-70.4= 12.6 84-70.4= 13.6
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01:31:42 ** x dev. from mean squared dev. 55 15.4 237.16 63 7.4 54.76 65 5.4 29.16 67 3.4 11.56 68 2.4 5.76 72 1.6 2.56 77 6.6 43.56 83 12.6 158.76 84 13.6 184.96 634 728.08 mean = 634 / 9 = 70.4 std. dev. = `sqrt (728.08 / 8) = 9.54 range = 84 - 55 = 29 **
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RESPONSE --> ok
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01:49:19 **** query probl 13.3.12 freq dist 14,8; 16,12; 18,15; 20,14; 22,10; 24,6; 26,3
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RESPONSE --> 6 4 3 6 12 18 23 mean-10.3 6-10.3=-4.3 18.49 4-10.3=-6.3 39.69 3-10.3=-7.3 53.29 6-10.3=-4.3 18.49 12-10.3=1.7 2.89 18-10.3=7.7 59.29 23-10.3=12.7 161.29
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01:49:33 ** Value freq Value * Freq Dev^2 * F 14 8 112 204.80 16 12 192 112.32 18 15 270 16.80 20 14 280 12.32 22 10 220 86.40 24 6 144 146.40 26 3 78 144.48 Total 68 1296 723.52 Total squared dev is 723.5 so ave squared dev is 723.5 / 68 = 10.6, approx. Std dev is sqrt(ave squared dev) = sqrt(10.6) = 3.3 approx. **
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RESPONSE --> ok
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01:54:14 **** query probl 13.3.18 chebyshev for z=5 What is the least possible number of elements of a sample which lie within 5 standard deviations of the mean?
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RESPONSE --> 1- 2/5= 0.60
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01:54:28 ** The formula 1 - 1/k^2 gave you .96. That's the proportion which must under any circumstances lie between mean and 5 std dev from the mean. So the number is .96 n, where n is the number of elements in the sample. **
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RESPONSE --> ok
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01:55:06 query probl 13.3.48 mean length of stay 2.7 days, std dev 7.1 days.
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RESPONSE -->
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01:55:12 ** A sketch of a normal distribution will be a normal, or bell-shaped curve with its peak at the mean, dropping to about 60% of peak value at 1 std dev from the mean and to about 14% of peak value at 2 std dev from the mean. For a normal distribution with mean 2.7 and std dev. 7.1 one std dev from the mean occurs at 2.7 + 7.1 = 9.8 and at 2.7 7.1 = -4.4; two std dev from the mean occurs at 16.9 and -11.5. The corresponding normal curve cannot represent length of stay, since length of stay must not be less than zero. As a result we obtain a curve which tails off for large values of x, but whose area is concentrated mostly between 0 and 2.7. This curve is not symmetric like the bell curve but is very skewed, bunched up on one side of the mean 2.7 and more spread out for larger values. **
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RESPONSE --> ok
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01:56:05 **** Describe your sketch of the distribution of lengths of stay.
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RESPONSE --> The bell shape would go from 2.7 days at the lowest and 7.1 days at the highest
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01:56:18 2.7 is in the center with each number within 7.1 of the right or left of 2.7 and each additonal number on the left or right within 7.1 of each other. I see the curve as not being skewed.
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RESPONSE --> ok
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01:57:10 **** Is your distribution skewed? If so why, and if not why do you think it shouldn't be?
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RESPONSE --> yes, because of the deviation of 7.1
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01:57:18 STUDENT RESPONSE AND INSTRUCTOR COMMENT: I said no, but I'm not really sure how to determine this if I draw it myself, becasue I place all the numbers at an equal distance from each other. It just hit me while typing this that it must be skewed to the right because I can't really have negative days stay in the hospital. I think I have confused myself ** You didn't confuse yourself. That's exactly the point. You can't stay fewer than 0 days; since even 1 standard deviation is way below 0 the deviations must be primarily to the right of the mean. So the distribution must be skewed significantly to the right. GENERAL SUGGESTION: In general to understand the graphs of various distributions, try to understand in terms first of the bell-shaped curve with max height at the mean, dropping to about 60% height at a distance of 1 std dev from the mean and to about 14% at 2 std dev from the mean. Then understand that this distribution can be distorted, or skewed, as in this problem. This occurs when most of the distribution lies close to the mean on one side, with a smaller part of the distribution spread out further from the mean on the other. **
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RESPONSE --> ok
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01:57:38 017. `query 17
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RESPONSE --> ok
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The probability of ending up 6 blocks South is therefore prob = C(10,8) * (1/2)^8 * (1/2)^2 = 45 * (1/2)^10 = 45 / 1024, or about .044. **
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RESPONSE --> ok
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00:16:09 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
.................................................ÕCֈw
assignment #001 001. Only assignment: prelim asst
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12:11:49 `q001. Part 1 includes six activities. If you have completed an activity, just enter the answer 'completed'. This question is appearing in the Question box. The box to the right is the Answer box, where you will type in your answers to the questions posed here. To use this program you read a question, then enter your answer in the Answer box and click on Enter Answer. In your answers give what is requested, but don't go into excruciating detail. Try to give just enough that the instructor can tell that you understand an item. After entering an answer click on Next Question/Answer above the Question box. Do you understand these instructions?
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RESPONSE --> yes, I understand these instructions
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12:14:42 This program has created the folder c:\vhmthphy on your hard drive. Browse to that folder and locate the file whose name begins with SEND. The name of this file will also include your name, as you gave it to the program, and the file will show as a Text file. Never tamper with a SEND file in any way. It contains internal codes as if these codes are tampered with you won't get credit for the assignment. However you are welcome to copy this file to another location and view it, make changes, etc. Just be sure that when requested to do so you send the instructor the original, tamper-free file. State in the Answer box whether or not you have been able to locate the SEND file. Don't send the SEND file yet. Note that more questions/instructions remain in the q_a_prelim.
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RESPONSE --> I have located the SEND file and I understand not to send this file.
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12:15:36 `q002. Note that every time you click on Enter Answer the program writes your response to your SEND file. Even if the program disappears all the information you have entered with the Enter Answer button will remain in that file. This program never 'unwrites' anything. Even if this program crashes your information will still be there in the SEND file. Explain this in your own words.
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RESPONSE --> Whatever I type in the enter response section, stays there. The computer will not lose this information, even if it were to crash.
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12:17:56 Any time you do not receive a reply from the instructor by the end of the following day, you should resubmit your work using the Resubmit Form at http://www.vhcc.edu/dsmith/genInfo/. You have already seen that page, but take another look at that page and be sure you see the Submit Work form, the Resubmit Form and a number of other forms that will be explained later. Enter a sentence or two describing the related links you see at that location.
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RESPONSE --> http://www.vhcc.edu/dsmith/genInfo/ is the site for resubmitting my work.
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12:18:41 `q003. If you are working on a VHCC computer, it is probably set up in such a way as to return to its original configuration when it is rebooted. To avoid losing information it is suggested that you back up your work frequently, either by emailing yourself a copy or by using a key drive or other device. This is a good idea on any computer. Please indicate your understanding of this suggestion.
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RESPONSE --> I understand that if I am working on a VHCC computer that I need to email myself a copy of my work.
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|lOؐг assignment #001
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12:20:40 `q001. Part 1 includes six activities. If you have completed an activity, just enter the answer 'completed'. This question is appearing in the Question box. The box to the right is the Answer box, where you will type in your answers to the questions posed here. To use this program you read a question, then enter your answer in the Answer box and click on Enter Answer. In your answers give what is requested, but don't go into excruciating detail. Try to give just enough that the instructor can tell that you understand an item. After entering an answer click on Next Question/Answer above the Question box. Do you understand these instructions?
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RESPONSE --> I understand
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12:21:08 This program has created the folder c:\vhmthphy on your hard drive. Browse to that folder and locate the file whose name begins with SEND. The name of this file will also include your name, as you gave it to the program, and the file will show as a Text file. Never tamper with a SEND file in any way. It contains internal codes as if these codes are tampered with you won't get credit for the assignment. However you are welcome to copy this file to another location and view it, make changes, etc. Just be sure that when requested to do so you send the instructor the original, tamper-free file. State in the Answer box whether or not you have been able to locate the SEND file. Don't send the SEND file yet. Note that more questions/instructions remain in the q_a_prelim.
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RESPONSE --> already completed
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12:21:22 `q002. Note that every time you click on Enter Answer the program writes your response to your SEND file. Even if the program disappears all the information you have entered with the Enter Answer button will remain in that file. This program never 'unwrites' anything. Even if this program crashes your information will still be there in the SEND file. Explain this in your own words.
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RESPONSE --> already completed
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12:21:30 Any time you do not receive a reply from the instructor by the end of the following day, you should resubmit your work using the Resubmit Form at http://www.vhcc.edu/dsmith/genInfo/. You have already seen that page, but take another look at that page and be sure you see the Submit Work form, the Resubmit Form and a number of other forms that will be explained later. Enter a sentence or two describing the related links you see at that location.
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RESPONSE --> already completed
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12:21:45 `q003. If you are working on a VHCC computer, it is probably set up in such a way as to return to its original configuration when it is rebooted. To avoid losing information it is suggested that you back up your work frequently, either by emailing yourself a copy or by using a key drive or other device. This is a good idea on any computer. Please indicate your understanding of this suggestion.
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RESPONSE --> already completed
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12:21:49 Once more, locate the SEND file in your c:\vhmthphy folder, and open the file. Copy its contents to the clipboard (this is a common operation, but in case you don't know how, just use CTRL-A to highlight the contents of the file and CTRL-C to copy the contents to the clipboard). Then return to the form that instructed you to run this program, and paste the contents into the indicated box (just right-click in the box and select Paste). You may now click on the Quit button, or simply close the program.
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RESPONSE --> already completed
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y mӄWxBō~ Student Name: assignment #004 004. Liberal Arts Mathematics
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12:23:34 `q001. Consider the statement 'If that group of six-year-olds doesn't have adult supervision, they won't act in an orderly manner.' Under which of the following circumstances would everyone have to agree that the statement is false? The group does have supervision and they do act in an orderly manner. The group doesn't have supervision and they don't act in an orderly manner. The group doesn't have supervision and they do act in an orderly manner. The group does have supervision and they don't act in an orderly manner.
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RESPONSE --> The group doesn't have supervision and they don't act in an orderly manner.
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12:23:54 The statement says that if the group doesn't have supervision, they will not act in an orderly manner. So if they don't have supervision and yet do act in an orderly manner the statement is contradicted. If the group does have supervision, the statement cannot be contradicted because condition of the statement, that the group doesn't have supervision, does not hold. The statement has nothing to say about what happens if the group does have supervision. Of course if the group doesn't have supervision and doesn't act in orderly manner this is completely consistent with the statement. Therefore the only way to statement can be considered false is the group doesn't have supervision and does act in an overly manner. Note that what we know, or think we know, about childrens' behavior has nothing at all to do with the logic of the situation. We could analyze the logic of a statement like 'If the Moon is made of green cheese then most six-year-olds prefer collard greens to chocolate ice cream'. Anything we know about the composition of the Moon or the tastes of children has nothing to do with the fact that the only way this statement could be shown false would be for the Moon to be made of green cheese and most six-year-olds to prefer the ice cream.
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RESPONSE --> ok
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12:27:37 `q002. List the different orders in which the letters a, b and c could be arranged (examples are 'acb' and 'cba'). Explain how you know that your list contains every possible order.
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RESPONSE --> abc, acb bac, bca cba, cab If you only use each letter once in each list, this is the possibilities
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12:28:01 The only reliable way to get all possible orders is to have a system which we are sure the list every order without missing any. Perhaps the simplest way to construct all possible orders is to list then alphabetically. We start with abc. There is only one other order that starts with a, and it is obtained by switching the last two letters to get acb. The next alphabetical order must start with b. The first possible listing starting with b must follow b with a, leaving c for last. The orders therefore bac. The only other order starting with b is bca. The next order must start with c, which will be followed by a to give us cab. The next order is obtained by switching the last two letters to get cba. This exhausts all possibilities for combinations of the three letters a, b and c. Our combinations are, in alphabetical order, abc, acb, bac, bca, cab, cba.
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RESPONSE --> ok
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12:31:04 `q003. One collection consists of the letters a, c, d and f. Another collection consists of the letters a, b, d and g. List the letters common to both collections. List the letters which appear in at least one of the collections. List the letters in the first half of the alphabet which do not appear in either of the collections.
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RESPONSE --> Letters common to both: a, d Letters in at least one collection: a, b, c, d, f, g Letters in the first half of the alphabet which do not appear in either collection: e
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12:31:46 To letters a and d each appear in both collections. No other letter does. The letters a, c, d, and f appear in the first collection, so they all in at least one of the collections. In addition to letters b and g appear in the second collection. Therefore letters a, b, c, d, f and g all appear in at least one of the collections. We consider the letters in the first half of the alphabet, in alphabetical order. a, b, c and d all appear in at least one of the collections, but the letter e does not. The letters f and g also appear in at least one of the collections, but none of the other letters of the alphabet do. The first half of the alphabet ends at m, so the list of letters in the first half of the alphabet which do not occur in at least one of the collections is e, h, i, j, k, l, m.
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RESPONSE --> ok
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12:34:30 `q004. Give the next element in each of the following patterns and explain how you obtained each: 2, 3, 5, 8, 12, ... 3, 6, 12, 24, ... 1, 3, 4, 7, 11, 18, ...
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RESPONSE --> 2, 3, 5, 8, 12, 17 3, 6, 12, 24, 48 1, 3, 4, 7, 11, 18, 29
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12:34:43 The pattern of the sequence 2, 3, 5, 8, 12, ... can be seen by subtracting each number from its successor. 3-2 = 1, 5-3 = 2, 8-5 = 3, 12-8 = 4. The sequence of differences is therefore 1, 2, 3, 4, ... . The next difference will be 5, indicating that the next number must be 12 + 5 = 17. The pattern of the sequence 3, 6, 12, 24, ... can be discovered by dividing each number into its successor. We obtain 6/3 = 2, 12/6 = 2, 24/12 = 2. This shows us that we are doubling each number to get the next. It follows that the next number in the sequence will be the double of 24, or 48. The pattern of the sequence 1, 3, 4, 7, 11, 18, ... is a little obvious. Starting with the third number in the sequence, each number is the sum of the two numbers proceeding. That is, 1 + 3 = 4, 3 + 4 = 7, 4 + 7 = 11, and 7 + 11 = 18. It follows that the next member should be 11 + 18 = 29.
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RESPONSE --> ok
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12:38:21 `q005. The number 18 can be 'broken down' into the product 9 * 2, which can then be broken down into the product 3 * 3 * 2, which cannot be broken down any further . Alternatively 18 could be broken down into 6 * 3, which can then be broken down into 2 * 3 * 3. Show how the numbers 28 and 34 can be broken down until they can't be broken down any further. Show that there at least two different ways to break down 28, but that when the breakdown is complete both ways end up giving you the same numbers.
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RESPONSE --> 28: 7*4, 7*2*2 34: 17*2
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12:38:33 A good system is to begin by attempting to divide the smallest possible number into the given number. In the case of 34 we see that the number can be divided by 2 give 34 = 2 * 17. It is clear that the factor 2 cannot be further broken down, and is easy to see that 17 cannot be further broken down. So the complete breakdown of 34 is 2 * 17. To breakdown 28 we can again divide by 2 to get 28 = 2 * 14. The number 2 cannot be further broken down, but 14 can be divided by 2 to give 14 = 2 * 7, which cannot be further broken down. Thus we have 28 = 2 * 2 * 7. The number 28 could also the broken down initially into 4 * 7. The 4 can be further broken down into 2 * 2, so again we get 28 = 2 * 2 * 7. It turns out that the breakdown of a given number always ends up with exactly same numbers, no matter what the initial breakdown.
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RESPONSE --> ok
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12:40:42 `q006. Give the average of the numbers in the following list: 3, 4, 6, 6, 7, 7, 9. By how much does each number differ from the average?
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RESPONSE --> The average is 6. The differences from this average are 3, 2, 0, 0, 1, 1, 3
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12:41:04 To average least 7 numbers we add them in divide by 7. We get a total of 3 + 4 + 6 + 6 + 7 + 7 + 9 = 42, which we then divide by 7 to get the average 42 / 7 = 6. We see that 3 differs from the average of 6 by 3, 4 differs from the average of 6 by 2, 6 differs from the average of 6 by 0, 7 differs from the average of 6 by 1, and 9 differs from the average of 6 by 3. A common error is to write the entire sequence of calculations on one line, as 3 + 4 + 6 + 6 + 7 + 7 + 9 = 42 / 7 = 6. This is a really terrible habit. The = sign indicates equality, and if one thing is equal to another, and this other today third thing, then the first thing must be equal to the third thing. This would mean that 3 + 4 + 6 + 6 + 7 + 7 + 9 would have to be equal to 6. This is clearly not the case. It is a serious error to use the = sign for anything but equality, and it should certainly not be used to indicate a sequence of calculations.
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RESPONSE --> ok
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12:42:33 `q007. Which of the following list of numbers is more spread out, 7, 8, 10, 10, 11, 13 or 894, 897, 902, 908, 910, 912? On what basis did you justify your answer?
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RESPONSE --> The larger numbers are more spread out than the smaller numbers. They differ less.
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12:42:52 The first set of numbers ranges from 7 to 13, a difference of only 6. The second set ranges from 894 to 912, a difference of 18. So it appears pretty clear that the second set has more variation the first. We might also look at the spacing between numbers, which in the first set is 1, 2, 0, 1, 2 and in the second set is 3, 5, 6, 2, 2. The spacing in the second set is clearly greater than the spacing in the first. There are other more sophisticated measures of the spread of a distribution of numbers, which you may encounter in your course.
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RESPONSE --> ok
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12:46:31 `q008. 12 is 9 more than 3 and also 4 times 3. We therefore say that 12 differs from 3 by 9, and that the ratio of 12 to 3 is 4. What is the ratio of 36 to 4 and by how much does 36 differ from 4? If 288 is in the same ratio to a certain number as 36 is to 4, what is that number?
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RESPONSE --> Ratio of 36 to 4 is 9. 36 differs from 4 by 32. Ratio of 288 to 36 is 32.
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12:46:58 Just as the ratio of 12 to 3 is 12 / 3 = 4, the ratio of 36 to 4 is 36 / 4 = 9. 36 differs from 4 by 36 - 4 = 32. Since the ratio of 36 to 4 is 9, the number 288 will be in the same ratio to a number which is 1/9 as great, or 288 / 9 = 32. Putting this another way, the question asks for a 'certain number', and 288 is in the same ratio to that number as 36 to 4. 36 is 9 times as great as 4, so 288 is 9 times as great as the desired number. The desired number is therefore 288/9 = 32.
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RESPONSE --> ok
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12:48:44 `q009. A triangle has sides 3, 4 and 5. Another triangle has the identical shape of the first but is larger. Its shorter sides are 12 and 16. What is the length of its longest side?
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RESPONSE --> the longest side would be 20
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12:48:56 ** You need to first see that that each side of the larger triangle is 4 times the length of the corresponding side of the smaller. This can be seen in many ways, one of the most reliable is to check out the short-side ratios, which are 12/3 = 4 and 16/4 = 4. Since we have a 4-to-1 ratio for each set of corresponding sides, the side of the larger triangle that corresponds to the side of length 5 is 4 * 5 = 20. **
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RESPONSE --> ok
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nEcwڌ撏 Student Name: assignment #001 001. typewriter notation
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12:52:53 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> x - 2 / x + 4 and (x - 2) / (x + 4) When you have parenthesis, you must get that answer first
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12:53:00 The order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction. It makes a big difference whether you subtract the 2 from the 2 or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract: 2 - 2 / 2 + 4 = 2 - 1 + 4 (do multiplications and divisions before additions and subtractions) = 5 (add and subtract in indicated order) If there are parentheses you evaluate the grouped expressions first: (x - 2) / (x - 4) = (2 - 2) / ( 4 - 2) = 0 / 2 = 0.
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RESPONSE --> ok
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12:58:26 `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2. Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.
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RESPONSE --> The first problem can be completed from left to right but the second has parenthesis, so that part must be completed first. x=2 2*2*2=8 8+4=12 2+4=8 2*2*2*2*2*2*2*2*2=512
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12:58:36 2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4. 2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power. If x = 2, then 2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8. and 2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.
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RESPONSE --> ok
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13:06:27 `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?
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RESPONSE --> x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x x = 2 2-3/ [ (2*2-5)^2*3*2+1] -2+7*2 2-3/ [(-1)^2*7]-2+14 2-3/[-1]-2+14 2-3/11 -1/11 -0.09
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13:06:44 The numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3. If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way. The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ] Evaluating the expression for x = 2: - 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 = 2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses 2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses 2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term; 2 - 3 / 7 - 2 + 14 = evaluate in brackets 13 4/7 or 95/7 or about 13.57 add and subtract in order. The details of the calculation 2 - 3 / 7 - 2 + 14: Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have 2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.
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RESPONSE --> ok
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13:11:57 `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
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RESPONSE --> x=4 (4-5)= -1 2x-1= 8-1=7 -1^7= 1 3/x-2= 3/4-2= 3/2 1+3/2= 4/2= 2
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13:12:56 We get (4-5)^2 * 4 - 1 + 3 / 1 - 4 = (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses = 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4 = 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get = 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4). COMMON ERROR: (4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 = -1 ^ 2*4 - 1 + 3 / 4-2 = -1 ^ 8 -1 + 3 / 4 - 2. INSTRUCTOR COMMENTS: There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first. Exponentiation precedes multiplication. Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4). Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power. -1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.
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RESPONSE --> ok
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Ӆx|LK̊ Student Name: assignment #002 002. Describing Graphs
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13:24:59 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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RESPONSE --> y=3x-4 y=3*-3-4 y=-9-4 y= -13 y=3x-4 y= 3*-2-4 y=-6-4 y= -10 y=3x-4 y=3*-1-4 y=-3-4 y= -7 y=3x-4 y=3*0-4 y=0-4 y= -4 y=3x-4 y=3*1-4 y=3-4 y= -1 y=3x-4 y=3*2-4 y=6-4 y= 2 y=3x-4 y=3*3-4 y=9-4 y= 5 (-3, -13) (-2, -10) (-1, -7) (0, -4) (1, -1) (2, 2) (3, 5)
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13:25:23 The graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.
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RESPONSE --> ok
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13:28:04 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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RESPONSE --> The steepness increases from left to right
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13:28:45 The graph forms a straight line with no change in steepness.
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RESPONSE --> ok
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13:30:11 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> 3/3, The slope rises 3 for every 3 numbers
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13:30:18 Between any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.
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RESPONSE --> ok ....
..... CpF`
assignment #003 cʓ齯̀q} li Liberal Arts Mathematics II 10-10-2006
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15:23:15 Query 11.3.20 5 prizes among 25 students
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RESPONSE --> ok
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15:24:36 ** There are 25 students available so there are 25 choices for the first student. On the second choice there are 24 students left so there would be 24 possibilities. Similarly on the third, fourth and fifth selections there would be 23, 22 and 21 choices. The result, by the Fundamental Counting Principle, is 25 * 24 * 23 * 22 * 21 choices. 25 * 24 * 23 * 22 * 21 = 25 ! / ( (25 - 5) !) since 25 ! / ( (25 - 5) !) = 25 ! / (20! * 5!) = 25 * 24 * 23 * 22 * 21 This is P(25, 5). We use permutations because order matters when there are 5 different prizes. **
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RESPONSE --> ok, sorry, I typed ok, too fast
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15:25:26 Is repetition allowed in this situation?
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RESPONSE --> no
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15:25:41 ** GOOD STUDENT ANSWER: no repetition is allowed because there are 5 different prizes, and you can't give the same one to two people **
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RESPONSE --> exactly
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15:29:27 Query 11.3.30 3-letter monogram all letters different
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RESPONSE --> N={A,B,C} {A,B,C}
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15:30:24 ** You are choosing 3 different letters, and since the monogram will be different if you change the order we can say that order definitely applies. If there is no restriction on any letter, other than the restriction of no repetitions, then there are 26 choices for the first letter, 25 for the second, 24 for the third so by the Fundamental Counting Principle there are 26 * 25 * 24 ordered choices. We can write this as P(26, 3), the number of possible permutations of 3 objects chosen without replacement from 26. P(26,3) = 26!/(26-3) ! = 26 * 25 * 24, in agreement with the previous expression. However the third initial must be the same as Judy's, which is `z'. Thus, since there can be no repetitions, there are only 25 possibilities for the first letter (can't be `z') and 24 for the second (can't be `z', can't be the first). So there are only 25 * 24 = 600 possibilities. **
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RESPONSE --> OK
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15:42:29 Query 11.3.42 divide 25 students into groups of 3,4,5,6,7. In how many ways can the students be grouped?
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RESPONSE --> C(N,R)=P(N,R)/R!=N!/R!(N-R)! C(25,3)=P(25,3)/3!=25!/3!(25-3)! 25= 25/66 0.378
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15:43:09 ** If we make the group of 3 first there are C(25, 3) possible choices. If we make the group of 4 next there are 22 students left from whom to choose so there are C(22, 4) possible choices. If we make the group of 5 next there are 18 students left from whom to choose so there are C(18, 5) possible choices. If we make the group of 6 next there are 13 students left from whom to choose so there are C(13, 6) possible choices. If we make the group of 7 next there are 7 students left from whom to choose so there are C(7, 7) possible choices. The Fundamental Counting Principle tells you that you have to multiply the number of ways of obtaining the first group by the number of ways of obtaining the second group by the number of ways of obtaining the 3rd group by the number of ways of obtaining the fourth group by the number of ways of obtaining the fifth group: C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7). Note that we could have chosen the groups in a different order, perhaps with the group of 7 first, the group of 6 second, etc.. The same reasoning would tell us that there are now C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) ways to do this. If the two expressions C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7) and C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) are both written down and simplified can get them both into the same form 25! / [ 3 ! * 4 ! * 5 ! * 6 ! * 7 ! ], which can then be further simplified by cancellation and then multiplied. **
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RESPONSE --> OK, I UNDERSTAND WHAT YOU WERE LOOKING FOR NOW
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15:50:08 Query 11.3.60 C(n,0)What is the value of C(n,0)?What is the value of C(8,0)?
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RESPONSE --> C(n,0)=0, N!=N-1 C(8,0)=40,319!
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15:51:39 ** This is equivalent to choosing 0 objects from n objects. No matter what n is, there is one way to do this, which is to choose nothing. As another example there are C(4,2) = 6 ways in which to obtain 2 Heads on four flips of a coin, C(4,3) = 4 ways to obtain 3 Heads, C(4,4) = 1 way to obtain 4 Heads. Obtaining 4 Heads is the same as obtaining 0 Tails, and of course C(4,0) tells you how many ways that are to obtain 0 Tails. So C(4,0) must be 1. The formula also gives us this: C(n, 0) = n ! / [ (n - 0) ! * 0 ! ] = n ! / ( n ! * 1) = 1. **
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RESPONSE --> OK, I'M CONFUSED. THIS PROBLEM DOESN'T MATCH UP WITH THE #60 IN THE BOOK
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秇i~~^}jD assignment #004 cʓ齯̀q} li Liberal Arts Mathematics II 10-10-2006
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15:54:59 Query 11.4.6 Find C(9,6) on Pascal's triangle.
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RESPONSE --> C(9,6) is located on row 9 of the triangle
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15:55:33 ** You need to go to the n=9 row, the r=6 position, which is the 10th row and the 7th number in the row. Looks like you went to the 6th position. Note that C(9,6) = 9! / ( 6! (6-3)! ) = 9*8*7 / (3*2*1) = 3 * 4 * 7 = 84 does agree with the number in the n = 9 row and the r = 6 position. Note also that since the first row is row 0 and the first element in every row is element 0, the n = 9 row is the 10th row, and the r = 6 position is the 7th number from the left. **
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RESPONSE --> I understand to start with row 0.
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15:59:52 Query 11.4.18 clueless check of four of nine possible classroomsHow many of the possible selections will fail to locate the classroom?
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RESPONSE --> 70 selections will fail
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16:00:20 ** There are C(9,4) possible combinations of four classrooms. There are 8 'incorrect' classrooms, so there are C(8, 4) ways for the check to yield a 'wrong' classroom. C(9,4) = 126 and C(8,4) = 70. Note that the chance of ending up in the right classroom is 56 / 126, a little less than 50-50, with 56 of the 126 possible ways being successful. **
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RESPONSE --> ok
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16:02:05 Query 11.4.30 What sequence by totaling diagonals of Pascal's Triangle?
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RESPONSE --> random walks
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16:04:33 ** The numbers 1, 1, 2, 3, 5, 8, 13, 21, ... form what is called the Fibonacci Sequence, which seems to occur in all sorts of unexpected places. The sums of the diagonals are all Fibonacci numbers. **
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RESPONSE --> This was something discussed in chapter 5, not chapter 11
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16:17:52 Query 11.4.42 (x+y)^8
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RESPONSE --> x^8+8x^7y+28x^6y^2+56x^5y^3+70x^4y^4+56x^3y^5+28x^2y^6+8xy^7+y^8
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16:18:30 **(x + y ) ^ 8 = x^8 + C(8,1) x^7 y + C(8,2) x^6 y^2 + C(8,3) x^5 y^3 + C(8,4) x^4 y^4 + C(8,5) x^3 y^5 + C(8,6) x^2 y^6 + C(8,7) x^7 y + y^8 = x^8 + 8 x^7 y + 28 x^6 y^2 + 56 x^5 y^3 + 70 x^4 y^4 + 56 x^3 y^5 + 28 x^2 y^6 + 8 x y^7 + y^8. **
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RESPONSE --> ok
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16:19:02 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> No comments
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16:20:29 ** STUDENT COMMENT: I was ok with this assignment until I got to problem 11.4 - 42 I do not understand the reasoning behind the following problems. 11.4 - 45: (2a + 5b)^4 = The binomial expansion is listed in the answer section, but I do not understand how they got there. INSTRUCTOR RESPONSE: Here is the solution for (2a + 5b) ^ 5. The application of the Binomial Theorem is clearer for the 5th power than the 4th; if you understand this you'll get the pattern for the 4th power. The answer is found from C(5,5) * (2a)^2 * (5b)^0 + C(5, 4) * (2a)^4 * (5b)^1 + C(5, 2) * (2a)^3 * (5b)^2 + . etc., following the pattern of the binomial expansion formula. Expanding the powers of 2a and 5b we get C(5, 5) * 32 a^5 + C(5, 4) * 16 a^4 * 5b + C(5, 3) * 8 a^3 * 25 b^2 + etc., which is equal to 1 * 32 a^5 + 5 * 16 a^4 * 5b + 10 * 8 a^3 * 25 b^2 + etc., or finally to 32 a^5 + 80 a^4 + 2000 a^3 + etc.. ANOTHER QUESTION: 11.4 -50 and 51 The rth or general term of the binomial expansion for (x = y)^n and (x + y)^14;5th term. INSTRUCTOR RESPONSE: The rth term of (x+y)^n will be C(n, r) * x^r * y^(n-r). You are choosing x from r of the binomials and y from the remaining n - r binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) . (x+y), where it is understood that we have (x+y) written n times. The 5th term of (x+y)^14 requires that you choose x from 5 of the binomials and y from the other 14-5 = 9 binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y). There are C(14, 5) ways to do this, and the result for every one of these ways is x^5 * y^14. So the 5th term is C(14,5) x^5 y^9. **
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RESPONSE --> ok, I finally got the hang of this assignment toward the end, only the last couple actually. I had a really hard time with this whole assignment. The book doesn't do a very good job with explaining these problems.
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{cϙys assignment #005 cʓ齯̀q} li Liberal Arts Mathematics II 10-10-2006
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17:33:08 Query 11.5.10 ways to get different number on each of two fair dice In how many ways can you get a different number on each of two fair dice? How would you obtain the answer to this question if you were marooned on a desert island with no book and no table?
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RESPONSE --> 72 ways, 36 chances for each di
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17:33:37 ** On two fair dice you have 6 possible outcomes on the first and 6 on the second. By the Fundamental Counting Principle there are therefore 6 * 6 = 36 possible outcomes. We can list these outcomes in the form of ordered pairs: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) Of these 36 outcomes there are six that have the same number on both dice. It follows that the remaining 3 - 6 = 30 have different numbers. So there are 30 ways to get different numbers on the two dice. Note that your chance of getting different numbers is therefore 30 / 36 = 5/6 = .8333... .**
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RESPONSE --> ok, I understand not to double and why
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17:34:52 Query 11.5.12 bridge hands more than one suit How many bridge hands contain more than one suit?
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RESPONSE --> 36
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17:35:53 ** There are 13 cards in a bridge hand. The number of possible bridge hands is therefore C(52, 13). There are 13 cards of each suit. The number of possible bridge hands with all cards in a given suit is therefore C(13, 13) = 1 (common sense is that there is only one way to get all 13 cards in a given suit, which is to get all the cards there are in that suit). Since there are 4 suits there are 4 * C(13, 13) = 4 * 1 = 4 possible one-suit hands. Of the number of hands having more than one suit is C(52, 13) - 4. **
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RESPONSE --> ok
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17:48:15 11.5.20 # subsets of 12-elt set with from 3 to 9 elts? How many subsets contain from three to nine elements and how did you obtain your answer (answer in detail)?
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RESPONSE --> 8^2=64 64-1=63 8 restaraunts total, only 2 serve seafood. Only 1of the selected restaraunts will be serving seafood.
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17:50:12 ** You need the number of subsets with 3 elements, with 4 elements, etc.. You will then add these numbers to get the total number of 3-, 4-, 5-, ., 9-element subsets. Start with a 3-e.ement subset. In a 12-element set, how many subsets have exactly three elements? You answer this by asking how many possibilities there are for the first element, then how many for the second, then how many for the third. You can choose the first element from the entire set of 12, so you have 12 choices. You have 11 elements from which to choose the second, so there are 11 choices. You then have 10 elements left from which to choose the third. So there are 12 * 11 * 10 ways to choose the elements. However, the order of a set doesn't matter. 3 elements could be ordered in 3! different ways, so there are 12 * 11 * 10 / 3! ways to choose different 3-element sets. This is equal to C(12,3). So there are C(12, 3) 3-elements subsets of a set of 12 elements. Reasoning similarly we find that there are C(12,4) ways to choose a 4-element subset. C(12,5) ways to choose a 5-element subset. C(12,6) ways to choose a 6-element subset. C(12,7) ways to choose a 7-element subset. C(12,8) ways to choose a 8-element subset. C(12,9) ways to choose a 9-element subset. We see that there are C(12,3) + C(12,4) + C(12,5) + C(12,6) + C(12,7) + C(12,8) + C(12,9) possible subsets with 3, 4, 5, 6, 7, 8 or 9 elements. Alternatively you can figure out how many sets have fewer than 3 or more than 9 elements. There are C(12, 0) + C(12, 1) + C(12, 2) = 1 + 12 + 66 = 79 sets with fewer than 3 elements, and C(12, 10) + C(12, 11) + C(12, 12) = 66 + 12 + 1 = 79 sets with more than 3. Since there are 2^12 = 4096 possible subsets of a 12-element set there are 4096 - 79 - 79 = 3938 sets with between 3 and 9 elements. **
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RESPONSE --> ok
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18:04:03 11.5.30 10200 ways to get a straight Verify that there are in fact 10200 ways to get a straight in a 5-card hand.
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RESPONSE --> C(52,5) 5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+52=1368 2,598,960-C(52,5)=2,597,592
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18:04:44 ** There are 9 choices for the denomination of the lowest card in a straight, which gives 36 cards that could be the low card. However but if aces can be high or low there are 40. There are then four choices for the next-higher card, four for the next after that, etc., giving 40*4*4*4*4 possibilities. **STUDENT COMMENT: I don 't understand this one . Idon't see where you get the 9 from. INSTRUCTOR RESPONSE: Cards run from 2 through 10, then the four face cards, then the ace. You need five consecutive cards to make a straight. The highest possible straight is therefore 10, Jack, Queen, King and Ace. The lowest is 2, 3, 4, 5, 6. The lowest card of the straight can be any number from 2 through 10. That is 9 possibilities. In some games the ace can be counted as the low card, below the 2, as well as the high card. In that case there would be one more possibility for a straight, which could then consists of denominations 1, 2, 3, 4, 5. *&*&
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RESPONSE --> ok, I don't understand
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18:17:39 11.5.36 3-digit #'s from {0, 1, ..., 6}; how many mult of 25?
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RESPONSE --> 3- digit #'s from {0,1,2,3,4,5,6} C(25,1)=25 C(25,2)=300
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18:20:14 ** A 3-digit number from the set has six choices for the first digit (can't start with 0) and 7 choices for each remaining digit. That makes 6 * 7 * 7 = 294 possibilities. A multiple of 25 is any number that ends with 00, 25, 50 or 75. SInce 7 isn't in the set you can't have 75, so there are three possibilities for the last two digits. There are six possible first digits, so from this set there are 6 * 3 = 18 possible 3-digit numbers which are multiples of 25. A listing would include 100, 125, 150, 200, 225, 250, 300, 325, 350, 400, 425, 450, 500, 525, 550, 600, 625, 650. Combinations aren't appropriate for two reasons. In the first place the uniformity criterion is not satisfied because different digits have different criteria (i.e., the first digit cannot be zero). In the second place we are not choosing object without replacement. The fundamental counting principle is the key here. STUDENT SOLUTION AND INSTRUCTOR RESPONSE: All I can come up with is C(7,2)=21. & choices of #s and the # must end in 0 or 5 making it 2 of the 7 choices INSTRUCTOR RESPONSE: Right reasoning on the individual coices but you're not choosing just any 3 of the 7 numbers (uniformity criterion isn't satisfied--second number has different criterion than first--so you wouldn't use permutations or combinations) and order does matter in any case so you wouldn't use combinations. You have 7 choices for the first and 2 for the second number so there are 7 * 2 = 14 multiples of 5. **
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RESPONSE --> I understand not to start with 0. Ok, I think I might understand now, after reading through your reasoning
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18:39:58 Query 11.5.48 # 3-digit counting #'s without digits 2,5,7,8?
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RESPONSE --> If you figure up 3-digit numbers that do not contain the digits 2,5,7,8, there are only 200 numbers left.
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18:40:36 ** there are 5 possible first digits (1, 3, 4, 6, or 9) and 6 possibilities for each of the last two digits. This gives you a total of 5 * 6 * 6 = 180 possibilities. **
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RESPONSE --> Ok, I still don't understand
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18:41:02 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I really had trouble with this section. I think I will try to get some assistance.
....... 㼯izUw[_Ȋjqy
assignment #006 cʓ齯̀q} li Liberal Arts Mathematics II 10-11-2006
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19:00:08 Query 12.1.6 8 girls 5 boys What is the probability that the first chosen is a girl?
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RESPONSE --> 13 people all together, 8 girls 8/13=0.62
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19:00:23 ** Assuming the choice is completely random there are 13 possible choices, 8 of which are female so we have P(female) = 8 / 13 = .6154, approx. **
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RESPONSE --> ok, I understand. I just rounded up
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19:05:30 Query 12.1.12 3 fair coins: Probability and odds of 3 Heads.
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RESPONSE --> 1/2 for each coin= 0.50 3/1=1/1 are the odds
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19:06:26 ** There are 8 equally likely possible outcomes when flipping 3 fair coins. You can list them: hhh, hht, hth, htt, thh, tht, tth, ttt. Or you can use the fact that there are 2 possibilities on each flip, therefore 2*2*2 = 2^3 = 8 possible outcomes. Only one of these outcomes, hhh, consists of 3 heads. The probability is therefore P(3 heads) = # of outcomes favorable/total number of possible outcomes = 1 / 8. The odds in favor of three heads are Odds ( 3 heads ) = # favorable to # unfavorable = 1 to 7. **
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RESPONSE --> ok
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19:12:14 Query 12.1.20 P(pink) from two pink parents (Rr and Rr) What is the probability of a pink offspring.
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RESPONSE --> If both parents are Pink, there should be a 100% chance of a pink offspring. However, Do the parents not still possess the red and white passed from their parents? If so, that is still s recessive trait and the chance is lowered to 25% to 75% depending on how you look at it.
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19:12:39 ** The genes R and r stand for the red and white genes. A pink offspring is either Rr or rR. RR will be red, rr white. R r R RR Rr r rR rr shows that {RR, Rr, rR, rr} is the set of equally likely outcomes. We season two of the four possible outcomes, rR and Rr, will be pink. So the probability of pink offspring is 2/4 = 1 / 2. **
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RESPONSE --> Ok, you are going by dominence. I understand
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19:15:42 Query 12.1.33 cystic fibrosis in 1 of 2K cauc, 1 in 250k noncauc What is the empirical probability, to 6 places, that a randomly chosen non-Caucasian newborn will have cystic fibrosis?
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RESPONSE --> 1/500=0.002 because this is based on the information given in problem 33. For CF, it is 0.000004
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19:15:48 ** There is 1 chance in 250,000 so the probability is 1 / 250,000 = 4 * 10^-6, or .000004. **
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RESPONSE --> ok
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19:18:07 12.1.40 Cc genes carrier, cc has disease; 2 carriers first child has disease **** What is the probability that the first child has the disease?
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RESPONSE --> If you write this out in table form, 1/4=0.25. This is the value for cc. There is also a cC so wouldn't they have it in some form as well, making it 0.50?
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19:18:37 ** If cc has the disease, then the probability that the first child will have the disease is 1/4. **
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RESPONSE --> ok, I see, it is only the obvious and I shouldn't look further.
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19:19:34 What is the sample space for this problem?
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RESPONSE --> I'm not sure what you are asking
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19:20:43 ** The sample space is {CC, Cc. cC, cc}. **
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RESPONSE --> oh, ok, that is what I meant by the table. Mine was the same and is below. C c C CC Cc c cC cc
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19:21:42 12.1.60 36 in class, 3 chosen **** What is the probability that the choice will be the given three people in any order?
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RESPONSE --> 3/36=0.083
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19:22:04 ** There are P(36,3) possible ordered choices of 3 people out of the 36. P(36,3) = 36! / (36-3)! = 36! / 33! = (36*35*34*33*32*31*...*1) / (33*32*31*...*1) = 36*35*34=40,000 or so. The probability of any given choice is therefore 1 / P(36,3) = 1/40,000 = .000025, approx.. For any given set of three people there are six possible orders in which they can be chosen. So the probability of the three given people, in any order, is 6 * probability of a given order = 6 / P(36,3) = 6/40,000 = .00015. Alternatively we can say that we are choosing 3 of 36 people without regard for order, so there are C(36,3) possibilities and the probability of any one of them is 1 / C(36,3). **
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RESPONSE --> ok
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19:26:02 Query 12.1.75 digits 1, 2, ..., 5 rand arranged; prob even, prob digits 1 and 5 even What is the probability that the resulting number is even and how did you obtain your answer?
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RESPONSE --> 2 and 4 are the even numbers in this problem 1+2+3+4+5=15 15/75=0.2
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19:26:40 ** The number will be even if it ends in 2 or 4. There are 5 possible ending numbers. So 2 of the 5 possible ending numbers are even and the probability of an even number is 2/5. We analyze in two ways the number of ways to choose a number with digits 1 and 5 even. First way: There are 5! = 120 possible arrangements of the 5 digits. There are only two possible even digits, from which we will choose digit 1 and digit 5. The order of our choice certainly matters, since a different choice will give us a different 5-digit number. So we are choosing 2 digits from a set of 2 digits, where order matters. We therefore have P(2, 2) = 2*1 / 0! = 2 ways to choose these digits. The remaining 3 digits will comprise digits 2, 3 and 4. We are therefore choosing 3 digits from a set of 3, in order. There are P(3, 3) = 3*2*1/0! = 6 ways to do so. To obtain our number we can choose digits 1 and 5, then digits 2, 3 and 4. There are P(2, 2) * P(3, 3) = 2 * 6 = 12 ways to do this. So the probability that digits 1 and 5 are even is 12 / 120 = 1/10. Second way: A simpler solution looks at just the possibilities for digits 1 and 5. There are P(2, 2) = 2 choices for which these digits are even, and P(5, 2) = 20 total choices for these two digits. The probability that both will be even is therefore 2/20 = 1/10, the same as before. **
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RESPONSE --> ok
...................................................... 㼯izUw[_Ȋjqy
assignment #006 cʓ齯̀q} li Liberal Arts Mathematics II 10-11-2006
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19:00:08 Query 12.1.6 8 girls 5 boys What is the probability that the first chosen is a girl?
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RESPONSE --> 13 people all together, 8 girls 8/13=0.62
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19:00:23 ** Assuming the choice is completely random there are 13 possible choices, 8 of which are female so we have P(female) = 8 / 13 = .6154, approx. **
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RESPONSE --> ok, I understand. I just rounded up
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19:05:30 Query 12.1.12 3 fair coins: Probability and odds of 3 Heads.
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RESPONSE --> 1/2 for each coin= 0.50 3/1=1/1 are the odds
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19:06:26 ** There are 8 equally likely possible outcomes when flipping 3 fair coins. You can list them: hhh, hht, hth, htt, thh, tht, tth, ttt. Or you can use the fact that there are 2 possibilities on each flip, therefore 2*2*2 = 2^3 = 8 possible outcomes. Only one of these outcomes, hhh, consists of 3 heads. The probability is therefore P(3 heads) = # of outcomes favorable/total number of possible outcomes = 1 / 8. The odds in favor of three heads are Odds ( 3 heads ) = # favorable to # unfavorable = 1 to 7. **
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RESPONSE --> ok
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19:12:14 Query 12.1.20 P(pink) from two pink parents (Rr and Rr) What is the probability of a pink offspring.
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RESPONSE --> If both parents are Pink, there should be a 100% chance of a pink offspring. However, Do the parents not still possess the red and white passed from their parents? If so, that is still s recessive trait and the chance is lowered to 25% to 75% depending on how you look at it.
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19:12:39 ** The genes R and r stand for the red and white genes. A pink offspring is either Rr or rR. RR will be red, rr white. R r R RR Rr r rR rr shows that {RR, Rr, rR, rr} is the set of equally likely outcomes. We season two of the four possible outcomes, rR and Rr, will be pink. So the probability of pink offspring is 2/4 = 1 / 2. **
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RESPONSE --> Ok, you are going by dominence. I understand
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19:15:42 Query 12.1.33 cystic fibrosis in 1 of 2K cauc, 1 in 250k noncauc What is the empirical probability, to 6 places, that a randomly chosen non-Caucasian newborn will have cystic fibrosis?
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RESPONSE --> 1/500=0.002 because this is based on the information given in problem 33. For CF, it is 0.000004
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19:15:48 ** There is 1 chance in 250,000 so the probability is 1 / 250,000 = 4 * 10^-6, or .000004. **
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RESPONSE --> ok
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19:18:07 12.1.40 Cc genes carrier, cc has disease; 2 carriers first child has disease **** What is the probability that the first child has the disease?
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RESPONSE --> If you write this out in table form, 1/4=0.25. This is the value for cc. There is also a cC so wouldn't they have it in some form as well, making it 0.50?
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19:18:37 ** If cc has the disease, then the probability that the first child will have the disease is 1/4. **
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RESPONSE --> ok, I see, it is only the obvious and I shouldn't look further.
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19:19:34 What is the sample space for this problem?
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RESPONSE --> I'm not sure what you are asking
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19:20:43 ** The sample space is {CC, Cc. cC, cc}. **
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RESPONSE --> oh, ok, that is what I meant by the table. Mine was the same and is below. C c C CC Cc c cC cc
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19:21:42 12.1.60 36 in class, 3 chosen **** What is the probability that the choice will be the given three people in any order?
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RESPONSE --> 3/36=0.083
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19:22:04 ** There are P(36,3) possible ordered choices of 3 people out of the 36. P(36,3) = 36! / (36-3)! = 36! / 33! = (36*35*34*33*32*31*...*1) / (33*32*31*...*1) = 36*35*34=40,000 or so. The probability of any given choice is therefore 1 / P(36,3) = 1/40,000 = .000025, approx.. For any given set of three people there are six possible orders in which they can be chosen. So the probability of the three given people, in any order, is 6 * probability of a given order = 6 / P(36,3) = 6/40,000 = .00015. Alternatively we can say that we are choosing 3 of 36 people without regard for order, so there are C(36,3) possibilities and the probability of any one of them is 1 / C(36,3). **
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RESPONSE --> ok
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19:26:02 Query 12.1.75 digits 1, 2, ..., 5 rand arranged; prob even, prob digits 1 and 5 even What is the probability that the resulting number is even and how did you obtain your answer?
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RESPONSE --> 2 and 4 are the even numbers in this problem 1+2+3+4+5=15 15/75=0.2
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19:26:40 ** The number will be even if it ends in 2 or 4. There are 5 possible ending numbers. So 2 of the 5 possible ending numbers are even and the probability of an even number is 2/5. We analyze in two ways the number of ways to choose a number with digits 1 and 5 even. First way: There are 5! = 120 possible arrangements of the 5 digits. There are only two possible even digits, from which we will choose digit 1 and digit 5. The order of our choice certainly matters, since a different choice will give us a different 5-digit number. So we are choosing 2 digits from a set of 2 digits, where order matters. We therefore have P(2, 2) = 2*1 / 0! = 2 ways to choose these digits. The remaining 3 digits will comprise digits 2, 3 and 4. We are therefore choosing 3 digits from a set of 3, in order. There are P(3, 3) = 3*2*1/0! = 6 ways to do so. To obtain our number we can choose digits 1 and 5, then digits 2, 3 and 4. There are P(2, 2) * P(3, 3) = 2 * 6 = 12 ways to do this. So the probability that digits 1 and 5 are even is 12 / 120 = 1/10. Second way: A simpler solution looks at just the possibilities for digits 1 and 5. There are P(2, 2) = 2 choices for which these digits are even, and P(5, 2) = 20 total choices for these two digits. The probability that both will be even is therefore 2/20 = 1/10, the same as before. **
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RESPONSE --> ok
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QinwϏ̧I assignment #007 cʓ齯̀q} li Liberal Arts Mathematics II 10-11-2006
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19:35:36 12.2.6 single die, p(odd or <5). What is the probability of getting an odd result or a result < 5?
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RESPONSE --> odd number 3/6=0.5 <5 4/6=0.67
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19:37:12 ** there are 3 possible odd outcomes and four outcomes less that 5 which would add up to 7 outcomes, except that 2 of the outcomes < 5 are alrealdy odd and won't be counted. Thus the number of outcomes which are odd or less that 5 is 3 + 4 - 2 = 5 (this expresses the rule that n(A U B) = n(A) + n(B) - n(A ^ B), where U and ^ stand for union and intersection, respectively ). Thus the probability is 5/6. In terms of the specific sample space: The sample space for the experiment is {1, 2, 3, 4, 5, 6}. Success corresponds to events in the subset {1, 2, 3, 4, 5}. There are 6 elements in the sample space, 5 in the subset consisting of successful outcomes. Thus the probability is 5/6. **
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RESPONSE --> ok
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19:38:55 Query 12.2.15 drawing neither heart nor 7 from full deck
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RESPONSE --> probability- 9/13=0.69 odds- 9 to 4
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19:39:08 ** The sample space consists of the 52 cards in a full deck. There are 39 cards that aren't hearts, four 7's but only three aren't hearts so there are 36 cards that aren't hearts or seven. The probability is therefore 36/52 = 9/13. The odds in favor of the event are 16 to 36 (number favorable to number unfavorable), which in reduced form is 4 to 9. **
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RESPONSE --> ok
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19:42:41 12.2.24 prob of black flush or two pairs
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RESPONSE --> black flush=5108/2598960=0.001965 two pairs=123552/2598960=0.047539
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19:43:27 ** There are C(13,5) = 1287 ways to get a flush in a given suit--gotta choose the 5 cards from the 13 cards in that suit. There are two black suits so there are 2 * 1287 = 2574 possible black flushes. As the text tells you there are 123,552 ways to get two pairs. You can incidentally get this as 13 * C(4, 2) * 12 * C(4, 2) * C(44, 1) / 2 (2 of the 4 cards in any of the 13 denominations, then 2 of the 4 cards in any of the remaining 12 denominations, divide by 2 because the two denominations could occur in any order, then 1 of the 44 remaining cards not in either of the two denominations. There is no way that a hand can be both a black flush and two pairs, so there is no overlap to worry about (i.e., n(A and B) = 0 so n(A or B) = n(A) + n(B) - n(A and B) = n(A) + n(B) ). Thus there are 123,552 + 2574 = 126,126 ways to get one or the other. The probability is therefore 126,126 / 2,598,960 = .0485, approx. **
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RESPONSE --> ok
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19:45:30 12.2.33 x is sum of 2-digit numbers from {1, 2, ..., 5}; prob dist for random vbl x
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RESPONSE --> x P(x) 3 0.1 4 0.1 5 0.2 6 0.2 7 0.2 8 0.1 9 0.1
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19:45:41 ** If 2 different numbers are chosen from the set (1, 2, 3, 4, 5} then the sum 3 can appear only as 1+2. 4 can appear only as 1+3, assuming numbers can't be repeated (so, for example, 2+2 is not allowed). 5 can occur as 1+4 or as 2+3. 6 can occur as 1+5 or as 2+3. 7 can occur as 2+5 or as 1+6. 8 can occur only as 3+5. 9 can occur only as 4+5. Of the 10 possible combinations, the sums 3, 4, 8 and 9 can occur only once each, so each has probability .1. The sums 5, 6 and 7 can occur 2 times each, so each has probability .2. The possible sums are as indicated in the table below. 1 2 3 4 5 1 3 4 5 6 2 5 6 7 3 7 8 4 9 This assumes selection without replacement. There are C(5, 2) = 10 possible outcomes, as can be verified by counting the outcomes in the table. 3, 4, 8 and 9 appear once each as outcomes, so each has probability 1/10. 5, 6 and 7 appear twice each as outcomes, so each has probability 2/10. x p(x) 3 .1 4 .1 5 .2 6 .2 7 .2 8 .1 9 .1 **
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RESPONSE --> ok
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19:52:12 Query 12.2.36 n(A)=a, n(S) = s; P(A')=? What is the P(A')?
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RESPONSE --> It is the probability that an event will occur within sample space S
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19:52:32 ** A' is everything that is not in A. There are a ways A can happen, and s possibilities in the sample space S, so there are s - a ways A' can happen. So of the s possibilities, s-a are in A'. Thus the probability of A' is P(A') = n(A') / n(S) = (s - a) / s. **
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RESPONSE --> ok
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19:55:50 Query 12.2.42 spinners with 1-4 and 8-10; prob product is even
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RESPONSE --> 3/4=0.75
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19:57:05 ** The first number can be 1, 2, 3 or 4. The second can be 8, 9 or 10. There are therefore 4 * 3 = 12 possible outcomes. The only way to get an odd outcome is for the two numbers to both be odd. There are only 2 ways that can happen (1 * 9 and 3 * 9). The other 10 products are all even. So the probability of an even number is 10 / 12 = 5/6 = .833... . Alternatively we can set up the sample space in the form of the table 8 9 10 1 8 9 10 2 16 18 20 3 24 27 30 4 32 36 40 We see directly from this sample space that 10 of the 12 possible outcomes are even. **
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RESPONSE --> ok
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e}dQ~§ assignment #008 cʓ齯̀q} li Liberal Arts Mathematics II 10-11-2006
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20:08:47 Query 12.3.6 two members chosen for committee, Republican or no. Are the two choices independent or dependent and why?
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RESPONSE --> They could be dependent because if the first member chosen was republican, more than likely, the second will also be republican.
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20:08:57 ** The choice of the second is influenced by the first choice. If a Republican is chosen on the first choice, then there are fewer Republicans available for the second choice and the probability of getting a Republican on the second choice is lower than if a Republican had not been chosen first. COMMON ERROR: they are independant because they were randomly selected....if it is random then one did not depend on the other. EXPLANATION: The selection was indeed random, but the makeup of the remaining group available on the second choice depends on the first choice. **
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RESPONSE --> ok
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20:23:00 Query 12.3.12 table of motivations by male, female What is the probability that an individual will be primarily motivated by money or creativity given that the individual is female?
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RESPONSE --> 14/32=0.438 money 13/34=0.382 creativity
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20:23:32 ** There are 13 primarily motivated by money and 14 by creativity. Each person can have only one primary motivation so there is no overlap between these two groups. There are thus 13 + 14 = 27 motivated by money or creativity, out of a total of 66 women, which gives probability 27 / 66 = 9/22 = .41 approx.. If M is the set motivated by money, C the set motivated by creativity and S the entire sample space then we have p ( M or C) = p(M) + p(C) - p(M and C) or in set notation p(M U C) = p(M) + p(C ) - p(M ^ C), with U and ^ standing for set union and intersection. Since M ^ C is empty, p(M ^ C) = 0. p(M) = n(M) / n(S) = 13 / 66 and p(C) = n(C) / n(S) = 14/66 so we have p (M U C) = 13/66 + 14/66 = 27/66 = 9/22 = .41 approx. **
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RESPONSE --> ok, I figured the wrong things
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20:30:23 Query 12.3.24 prob of club 2d given diamond first What is the desired probability and how did you obtain it?
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RESPONSE --> There are four sets, club, diamond, spades, and hearts. If a diamond is dealt first, and club second, this probability is 1/4=0.25
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20:31:06 ** The probability is 13 / 51. When the second card is chosen there are 13 clubs still left, out of 51 remaining cards. This can also be calculated using P(A|B) = P(A^B) / P(B). The probability of getting a diamond on the first card and a club on the second is 13/52 * 13/51. The probability of getting a diamond on the first card is 13/52. So the probability of a club on the second given a diamond on the first is (13 / 52 * 13 / 51) / (13/52) = 13/51. **
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RESPONSE --> ok
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20:32:20 Query 12.3.32 prob of diamond given red What is the probability of getting a diamond given that the card is red?
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RESPONSE --> Well, it could either be a heart or a diamond, so it is 1/2, 0.50
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20:32:29 ** Of the 26 red cards, 13 are diamonds. So the probability of a diamond, given red, is 13/26 = 1/2. **
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RESPONSE --> ok
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20:38:34 Query 12.3.36 P(sale > $100) = .8; prob that first three sales all >$100 What is the probability that the first three sales are all for > $100 and how did you obtain your result?
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RESPONSE --> 0.3 you asked only for thee probability that all 3 sales would be >$100. That is 1/3=0.3
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20:38:49 ** the first sale has to be > $100, AND the second sale has to be > $100, AND the third sale has to be > $100. The events are independent. So the total probability, by the fundamental counting principle, is .8 * .8 * .8 = .512. **
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RESPONSE --> ok
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20:40:11 Query 12.3.42 P(critical direction) = .05; prob that none of the 5 scheduled days for launches has cloud movement in critical direction
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RESPONSE --> 0.015= 0.5^5
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20:40:25 ** On a given day cloud movement is not in critical direction with probability 1 - .05 = .95. this has to occur on the first day, then it has to occur on the second day, then on the third, then on the fourth, then on the fifth. These events are considered independent so the probability is .95 * .95 * .95 * .95 * .95 = .774 approx. (use your calculator to get the accurate answer). In order for none of the five days to have cloud movement in the critical direction, each of the five days must not have movement in the critical direction. The probability that the movement will not the in the critical direction for each of the days is .95. The probability that this will happen on every one of the five days is therefore .95 * .95 * .95 * .95 * .95 = .774, approx. **
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RESPONSE --> ok, I did it backward
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20:42:29 Query 12.3.54 probability of heads .52, tails .48; P(ht) What is the probability of head then tails?
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RESPONSE --> It is either 0.04 by finding the difference or 1
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20:42:44 ** There is a .52 probability of getting heads, then there is a .48 probability a getting tails. The two events have to happen consecutively. By the Fundamental Counting Principle there is thus a probability of .52 * .48 = .2496 of getting Heads then Tails. **
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RESPONSE --> ok
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20:45:17 Query 12.3.66 for given day P(rain)=.5, P(rain | rain day before) = .8, P(rain | no rain day before ) = .3. find P(rain on 3 consecutive days). For first 4 days in November what is the probability that it will rain on all four days given Oct 31 is clear?
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RESPONSE --> 0.04=0.5*0.8*0.3
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20:46:11 ** The probability of rain on the first of the four days is .3, since it is given that there was no rain on the previous day. The probability of rain on each of the following for days is .8, since on each of these days it rained the day before. The probability of rain on all four days is therefore .3 * .8 * .8 * .8 = .154. ANOTHER WAY OF SAYING IT: Oct 31 was clear so the probability of rain on the first day is .3. If it rained on the first day of the month then there is a probability of .8 that it rains on the second day. If it rained on the second day of the month then there is a probability of .8 that it rains on the third day. If it rained on the third day of the month then there is a probability of .8 that it rains on the fourth day. So the probability of rain on all 4 days is .3 * .8 * .8 * .8 = .154 **
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RESPONSE --> ok, I see how you got that. I needed to use the information given that 0.8 was the figure given for rain X 2 days
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20:47:19 Query 12.3.66 for given day P(rain)=.5, P(rain | rain day before) = .8, P(rain | no rain day before ) = .3. What is P(rain on 3 consecutive days).
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RESPONSE --> 0.8*0.3=0.24
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20:48:00 ** To get rain on 3 consecutive days requires rain on the first day, which happens with probability .5; then rain on the second day given that there was rain on the first day, which is .8, then rain on the third day, given that there was rain on the previous day; this third probability is also .8. The probability of the 3 events all happening (rain of 1st day AND rain on the second day AND rain on the third day) is therefore .5 * .8 * .8 = .32. **
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RESPONSE --> Ok, Why would you not use the day that rain had occurred but not the day before?
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20:48:09 QUESTION ON PROBLEM 33: Please explain Problem 33 of 12.3. It reads: If one number is chosen randomly from the intergers 1 throught 10, the probability of getting a number that is odd and prime, by the general multiplication rule is P(odd) * P(prime/odd) = 5/10 * 3/5 = 3/10 My question is how did we get three prime numbers out of 1 through 10? I assumed there were 4 of them (2, 3, 5, and 7).
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RESPONSE --> ok
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20:48:16 ** ONE ANSWER: The sample space is reduced to odd numbers, and 2 is not odd. So the set within the restricted sample space {1, 3, 5, 7, 9} should just be {3, 5, 7}. ANOTHER ANSWER: If we don't use the restricted sample space then we have P(prime | odd ) = P(prme) * P(odd | prime). We find P(prime) and P(odd | prime). P(prime) = 4 / 10, since there are 4 primes between 1 and 10. Within the unrestricted sample space P(odd | prime) is 3 / 4 since of the primes 2, 3, 5, 7 only three are odd. }Thus when you multiply P(prme) * P(odd | prime) you get 4/10 * 3/4 = 3/10, just as before. **
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RESPONSE --> ok
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20:48:44 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I didn't do as well with this section but I think with a little more practice, I should be fine.
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QP~GݴS{° assignment #009 cʓ齯̀q} li Liberal Arts Mathematics II 10-11-2006
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20:52:00 Query 12.4.3 P(2 H on 3 flips)
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RESPONSE --> 2/3*2/3*2/3=8/27=0.30
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20:52:31 ** On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of there 8 possibilities, only 3 of them have two Heads. Thus the probability is 3 / 8. You can get this result without listing. There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. If you have n flips, there are C(n,r) ways to get r Heads. This number appears in the n+1 row, as the r+1 entry, of Pascal's triangle. **
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RESPONSE --> ok
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20:53:52 What is the significance of .5^2 * .5 for this question?
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RESPONSE --> I'm not sure how this goes into the previous question
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20:54:54 ** .5^2 is the probability of getting Heads twice in a row. .5 is the probability of a Tails. .5^2 * .5 is therefore the probability of getting HHT. Since the probabilities are independent, you have the same probability of getting two Heads and a Tail in some different order. Since there are C(3,2) possible orders for 2 Heads on 3 coins, the probability of getting 2 Heads and one Tail is C(3,2) * .5^2 * .5 = 3 * .125 = .375, the same as the 3/8 we obtained by listing. **
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RESPONSE --> ok
.................... ִՉoDMj`
assignment #009 cʓ齯̀q} li Liberal Arts Mathematics II 11-27-2006
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23:59:08 Query 12.4.3 P(2 H on 3 flips)
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RESPONSE --> 3/8=0.375
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23:59:19 ** On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of there 8 possibilities, only 3 of them have two Heads. Thus the probability is 3 / 8. You can get this result without listing. There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. If you have n flips, there are C(n,r) ways to get r Heads. This number appears in the n+1 row, as the r+1 entry, of Pascal's triangle. **
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RESPONSE --> ok
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00:01:52 What is the significance of .5^2 * .5 for this question?
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RESPONSE --> it is the probability of getting heads twice in a row
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00:01:59 ** .5^2 is the probability of getting Heads twice in a row. .5 is the probability of a Tails. .5^2 * .5 is therefore the probability of getting HHT. Since the probabilities are independent, you have the same probability of getting two Heads and a Tail in some different order. Since there are C(3,2) possible orders for 2 Heads on 3 coins, the probability of getting 2 Heads and one Tail is C(3,2) * .5^2 * .5 = 3 * .125 = .375, the same as the 3/8 we obtained by listing. **
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RESPONSE --> ok
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00:03:43 Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> 7/8= 0.875 this is the probability of getting no heads on three flips
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00:03:51 ** Probability of getting no heads on three flips is P(TTT) = .5 * .5 * .5 = .125, or 1/8, obtained by multiplying the probability of getting a tails for each of 3 independent flips. Subtracting this from 1 gives .875, or 7/8. **
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RESPONSE --> ok
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00:06:22 Query 12.4.15 P(3 H on 7 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> c(7,3)=7*6*5/3= 35 ways 35/128
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00:06:28 ** There are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to choose three of the 7 `positions' for Heads on 7 flips. So there are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to get three heads on 7 flips. The probability of any of these ways is (1/2)^3 * (1/2)^4 = 1 / 2^7 = 1 / 128. The probability of 3 Heads on 7 flips is therefore 35 * 1/128 = 35 / 128. **
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RESPONSE --> ok
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00:08:39 Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die
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RESPONSE --> c(3,1)*1/6*(5,6)^2=4*1/6*25/36=100/216=25/72
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00:08:43 ** To get 1 success on 3 tries you have to get 1 success and 2 failures. On any flip the probability of success is 1/6 and the probability of failure is 5/6. For any ordered sequence with 1 success and 2 failures the probability is 1/6 * (5/6)^2. Since there are C(3,1) = 3 possible orders in which exactly 1 success can be obtained, the probability is C(3,1) * 1/6 * (5/6)^2 = 4 * 1/6 * 25 / 36 = 100 / 216 = 25 / 72. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/6, prob of failure q = 1 - 1/6 = 5/6, n = 3 trials and r = 1 success. **
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RESPONSE --> ok
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00:10:52 Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability?
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RESPONSE --> 1/3 (1/3)^7*(2/3)^3 c(10,7) 320/19683=0.016
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00:10:56 ** The probability of a correct answer from a random choice on any single question is 1/3. For any sequence of 7 correct answers and 3 incorrect the probability is (1/3)^7 * (2/3)^3. There are C(10,7) possible positions for 7 correct answers among 10 questions. So the probability is C(10,7) * (1/3)^7 * (2/3)^3 = 320/19683 = 0.0163 approx. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/3, prob of failure q = 1 - 1/3 = 2/3, n = 10 trials and r = 7 success. ANOTHER SOLUTION: There are C(10,7) ways to distribute the 7 correct answers among the 10 questions. The probability of any single outcome with 7 successes and 3 failures is the product of (1/3)^7, representing 7 successes, and (2/3)^3, representing 3 failures. The probability of exactly seven correct questions is therefore prob = C(10,7) * (2/3)^3 * (1/3)^7 . **
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RESPONSE --> ok
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00:12:11 Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each
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RESPONSE --> 0.448
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00:12:18 ** The probability of 0 side effects is C(8,0) * .7^8. The probability of 1 side effect is C(8,1) * .7^7 * .3^1. The probability of 2 side effects is C(8,2) * .7^6 * .3^2. The sum of these two probabilities is the probability that two or fewer patients will have side effects. We subtract this probability from 1 to get the probability that more than 2 will experience side effects. The result is approximately .448. DER**
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RESPONSE --> ok
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00:13:10 Query 12.4.48 P(4 th child is 1 st daughter) What is the probability that the fourth child is the first daughter and how did you obtain your result?
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RESPONSE --> 1/2*1/2*1/2*1/2=0.0625
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00:13:16 ** The fourth child will be the first daughter if the sequence is SSSD, S standing for son and D for daughter. The probability of S on any birth is .5, and the probability of G is .5. The probability of SSSD is .5^3 * .5 = .0625 or 1/16. **
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RESPONSE --> ok
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00:15:50 Query 12.4.54 10-step rnd walk, 1 dim; P(6 South) What is the probability of ending up 6 blocks South of the starting point and how did you obtain it?
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RESPONSE --> 1/10*1/6=0.016
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00:15:57 ** To end up 6 blocks South requires 8 steps South and 2 steps North. Thus exactly 8 of the 10 steps must be South, and there are C(10,8) ways for this to happen. The probability of any given combination of 8 South and 2 North is (1/2)^8 * (1/2)^2 = 1 / 2^10 = 1 / 1024. The probability of ending up 6 blocks South is therefore prob = C(10,8) * (1/2)^8 * (1/2)^2 = 45 * (1/2)^10 = 45 / 1024, or about .044. **
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RESPONSE --> ok
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00:16:09 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
........................................................................................................................................................... خ܀Jé{җ˨[X͂
assignment #010 cʓ齯̀q} li Liberal Arts Mathematics II 11-28-2006
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00:24:16 Query 12.5.6 fair dice game pays $3 for 6, $2 for 5, $1 for 4. What is a fair price to pay for playing this game?
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RESPONSE --> 3.00 per game
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00:24:30 ** A 1 in 6 chance of getting $3 is worth 1/6 * $3 = $.50 . A 1 in 6 chance of getting $2 is worth 1/6 * $2 = $.33 1/3 . A 1 in 6 chance of getting $1 is worth 1/6 * $1 = $.16 2/3 . The total expectation is $1.00 * 1/6 + $2.00 * 1/6 + $3.00 * 1/6 = $1.00 So a fair price to pay is $1.00 **
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RESPONSE --> ok
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00:28:00 Query 12.5.10 expectation Roulette $1 bet 18 red, 18 black one zero What is the expected net value of a bet on red?
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RESPONSE --> ($1)18/37=0.0486
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00:28:13 ** If your net gain is $1 for a win and -$1 for a loss the expected value is 18/37 * (+1) + 19/37 * (-1) = -$.027. **
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RESPONSE --> ok
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00:30:07 Query 12.5.20 exp sum of 2 of 5 cards 1-5. What is the expected sum of the numbers on the two cards drawn?
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RESPONSE --> 1/5+1/5=2/5
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00:30:17 ** You can't get a sum of 1 on two cards. There is also no way to get a sum of two, since the lowest total possible is 1 + 2 = 3. There are 2 ways to get total 3. You can get 1 on the first and 2 on the second, or vice versa. There are 2 ways to get total 4. You can get 1 on the first and 3 on the second, or vice versa. There are 4 ways to get total 5. You can get 1 on the first and 4 on the second, or vice versa, or 2 on the first and 3 on the second, or vice versa. There are 4 ways to get total 6. You can get 1 on the first and 5 on the second, or vice versa, or 2 on the first and 4 on the second, or vice versa. There are 4 ways to get total 7. You can get 2 on the first and 5 on the second, or vice versa, or 4 on the first and 3 on the second, or vice versa. There are 2 ways to get total 8. You can get 3 on the first and 5 on the second, or vice versa. There are 2 ways to get total 9. You can get 4 on the first and 5 on the second, or vice versa. You can't get more than 9. There are 2+2+4+4+4+2+2 = 20 possibilities, so the probabilities are 2/20, 4/20, 5/20, etc.. The expected sum is therefore 2/20 * 3 + 2/20 * 4 + 4/20 * 5 + 4/20 * 6 + 4/20 * 7 + 2/20 * 8 + 2/20 * 9. This gives 120 / 20 = 6. **
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RESPONSE --> ok
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00:30:23 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ok
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Ƶw׆wǪ assignment #011 cʓ齯̀q} li Liberal Arts Mathematics II 11-28-2006
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00:38:55 **** Query 12.6.6 rnd # table to simulate 50 one-and-one foul shooting opportunities if 70% prob of success; 2 shots Give the results of your tally. How does your empirical probability compare with the theoretical probability?
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RESPONSE --> Pts. Tally 0 2 1 0 2 0
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00:39:10 ** In 1-and-1 shooting you only get a second shot if you make the first. So there are 3 possibilities: miss the first, don't get another shot make the first, get another shot and make it make the first, get another shot and miss it prob of 0 = prob of miss on first shot = .3 prob of 1 = prob of hit on first and miss on 2d = .3 * .7 = .21 prob of 2 = prob ot hit * prob of hit = .49. 'Hits' happen with 70% or .7 probability, misses with probability 30% or 3. The theoretical probability of 2 misses is probability of miss * probability of miss = .30 * .30 = .09. The theoretical probability of 2 miss and 1 hit is probability of miss * probability of hit + probability of miss *hit probability of miss = .30 * .70 + .70 * .30 = .21 + .21 = .42. The theoretical probability of 2 hits is probability of hit * probability of hit = .70 * .70 = .49. Note that these probabilities add up to .09 + .42 + .49 = 1, as they must since these three events cover all possibilities. To use the table, randomly pick a starting point. Let numbers 1-7 correspond to making the free throw, with 8, 9 and 0 corresponding to misses. Go down the list, or across the list in an order you decided before looking at the list. Read two digits from the list and see if they correspond to two 'hits', two 'misses' or a 'hit' and a 'miss'. Record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Read two more digits and record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Continue until you have the required number of results. Tally how many times you got 0 'hits', 1 'hit', 2 'hits' etc.. Any outcome that starts with a 'miss' corresponds to zero point. 'Hit-miss' corresopnds to 1 point and 'hit-hit' corresponds to 2 points. Determine the percent of time you got each number of points, and compare to the theoretical probabilities .09, .42 and .49. *&*& **
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RESPONSE --> ok
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00:43:40 Query 12.6.12 rnd walk start N then right, left or straight with prob 1/2, 1/6, 1/3; 1 st 2 columns of table
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RESPONSE --> 1/2*1/6*1/3=0.02
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00:45:12 ** Your probabilities are given as 1/2, 1/6 and 1/3. These can all be expressed in terms of the common denominator 6: 1/2 = 3/6, 1/6 = 1/6, 1/3 = 2/6. So a move to the right has 3 chances out of 6, a move to the left has 1 chance out of 6 and a move straight has 2 chances out of 6. You can simulate this by letting the three digits 1, 2, 3 stand for a move to the right, the single digit 4 for a move to the left and the two digits 5, 6 for a straight move. The remaining digits 0, 7, 8, 9 don't stand for anything, and if you land on one of these numbers you just move to the next number. So according to your the first two columns of you table, how many times do you move to the right, how many to the left, how many straight and where do you end up? **
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RESPONSE --> 4 lefts, 4 straights back at the beginning?
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00:45:24 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I'm completely confused
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00:45:34 014. `query 14
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RESPONSE --> ok
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00:52:11 query problem 13.1.6 freq dist 35 IQ scores class width 5
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RESPONSE --> I'm not sure if I understand this but if I did it correctly on paper, the graph shows an inrease in IQ scores, starting at 91 and going all the way to 134
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00:52:53 ** These numbers might or might not be completely accurate, but the calculation of the relative frequencies as percents follows from the tallies: 91-95 1 1/50 = 2% 96-100 3 3/50 = 6% 101-105 5 5/50 = 10% 106-110 7 7/50 = 14% 111-115 12 12/50 = 24% 116-120 9 9/50 = 18% 121-125 8 8/50 = 16% 126-130 3 3/50 = 6% 131-135 2 2/50 = 4% **
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RESPONSE --> oh, I see how it goes now
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01:04:26 query problem 13.1.10 stem and leaf for yards gained by 44 rushers
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RESPONSE --> 0 3,7 1 2,6,8,9 2 2,4,5,8,9 3 0,2,3,3,6,6,7,9 4 0,1,2,3,3,5,6,9 5 1,4,5,8 6 0,2,7 7 3,3,9 8 6,8 9 4 10 2 11 2 12 3
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01:04:40 ** In order we get the following: 0 3 7 1 2 6 8 9 2 2 4 5 8 9 9 3 0 2 3 3 6 6 7 9 4 0 1 2 3 3 5 6 9 5 1 4 5 8 6 0 2 7 7 3 3 9 8 6 8 9 4 10 2 11 2 12 3 **
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RESPONSE --> ok
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01:06:26 query problem 13.1.35 empirical probability distribution for letters of alphabet
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RESPONSE --> letter probability A 0.208 E 0.338 I 0.169 O 0.208 U 0.078
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01:06:55 ** The probabilities for A, E, I, O, U are: .08. .13, .065, .08, .03. The sum of these probabilities is .385 The probabilities of the letters given a vowel: .08 / .385 = .208 .13 / .385 = .338 .065 / .385 = .169 .08 / .385 = .208 .03 / .385 = .078 These probabilities total 1. **
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RESPONSE --> OK
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01:07:05 015. `query 15
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RESPONSE --> OK
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01:11:29 query problem 13.2.10 .3, .4, .3, .8, .7, .9, .2, .1, .5, .9, .6
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RESPONSE --> MEAN- 0.52 MEDIAN- 0.5 MODE- 0.3, 0.9
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01:11:54 ** The numbers, in order, are .1, .2, .3, .3, .4, .5, .6, .7, .8, .9, .9 The mean, obtained by adding the 11 numbers then dividing by 11, is .518. The median occurs at position (n + 1 ) / 2 = 6 in the ordered list. This number is .5. Note that there are five numbers before .5 and five numbers after .5. The maximum number of times a number repeats in this distribution is 2. So there are two modes (and we say that the distribution is bimodal). The modes are .3 and .9. **
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RESPONSE --> I miscounted
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01:16:50 **** query problem 13.2.24 more effect from extreme value
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RESPONSE --> the mean would show more of an effect
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01:17:01 ** The mean is drastically affected by the error; correcting the error changes the mean by about 3 units. The median number, however, simply shifts 1 position, changing from 2.28 to 2.39. **
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RESPONSE --> ok
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01:21:43 **** query problem 13.2.30 Salaries 6 @$19k, 8 @ 23k, 2 @ 34.5k, 7 @ 56.9k, 1 @ 145.5k.
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RESPONSE --> the mean is 135,000
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01:22:03 ** IF THERE ARE 28 EMPLOYEES: The totals paid for each salary level are: 6 * $19,500 = $117,000 8 * $23,000 = $184,000 4 * $28,300 = $113,200 2 * $34,500 = $69,000 7 * $36,900 = $258,300 1 * $145,500 = $145,500 The grand total paid in salaries to the 28 employees is therefore $887,000, giving an average of $887,000 / 28 = $31,700. The median occurs at position (n + 1) / 2 = (28 + 1) / 2 = 14.5. Since the 14 th salaray on a list ordered from least to greatest is $23,000 and the 15 th is $28300 the median is ($23000 +$28300) / 2 = $25,650. The mode is 23,000, since this salary occurs more frequently than any other. IF THERE ARE 24 EMPLOYEES: The totals paid for each salary level are: $19,000 * 6 = $114,000 $23,000 * 8 = $184,000 $34,500 * 2 = $69,000 $56,900 * 7 = $398,300 $145,500 * 1 = $145,500 Adding these gives a grand total, which is divided by the number 24 of employees to obtain the mean $37,950. The median occurs at position (n + 1) / 2 = (24 + 1) / 2 = 12.5. Since the $23000 salary covers positions 7 thru 14 in an ordered lise of salaries the median is $23,000. The mode is 23,000, since this salary occurs more frequently than any other.
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RESPONSE --> ok
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01:22:56 **** query problem 13.2.51 mean, med, mode of 0, 1, 3, 14, 14, 15, 16, 16, 17, 17, 18, 18, 18, 19, 20
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RESPONSE --> mean- 13.7 median- 16 mode- 18
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01:23:02 ** The mean is 13.73, obtained by adding together all the numbers and dividing by n = 15. The median is in position (n+1) / 2 = (15+1)/2 = 8 on the ordered list; the 8 th number is 16. The mode is 18, which is the only number occurring as many as 3 times. **
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RESPONSE --> ok
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01:23:07 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ok
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01:23:24 016. `query 16
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RESPONSE --> ok
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01:31:18 query probl 13.3.6 range, std dev of {67, 83, 55, 68, 77, 63, 84, 72, 65}
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RESPONSE --> range-84-55=29 mean- 70.4 standard deviation- 55-70.4= -15.4 63-70.4= -7.4 65-70.4= -5.4 67-70.4= -3.4 68-70.4= -2.4 72-70.4= 1.6 77-70.4= 6.6 83-70.4= 12.6 84-70.4= 13.6
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01:31:42 ** x dev. from mean squared dev. 55 15.4 237.16 63 7.4 54.76 65 5.4 29.16 67 3.4 11.56 68 2.4 5.76 72 1.6 2.56 77 6.6 43.56 83 12.6 158.76 84 13.6 184.96 634 728.08 mean = 634 / 9 = 70.4 std. dev. = `sqrt (728.08 / 8) = 9.54 range = 84 - 55 = 29 **
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RESPONSE --> ok
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01:49:19 **** query probl 13.3.12 freq dist 14,8; 16,12; 18,15; 20,14; 22,10; 24,6; 26,3
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RESPONSE --> 6 4 3 6 12 18 23 mean-10.3 6-10.3=-4.3 18.49 4-10.3=-6.3 39.69 3-10.3=-7.3 53.29 6-10.3=-4.3 18.49 12-10.3=1.7 2.89 18-10.3=7.7 59.29 23-10.3=12.7 161.29
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01:49:33 ** Value freq Value * Freq Dev^2 * F 14 8 112 204.80 16 12 192 112.32 18 15 270 16.80 20 14 280 12.32 22 10 220 86.40 24 6 144 146.40 26 3 78 144.48 Total 68 1296 723.52 Total squared dev is 723.5 so ave squared dev is 723.5 / 68 = 10.6, approx. Std dev is sqrt(ave squared dev) = sqrt(10.6) = 3.3 approx. **
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RESPONSE --> ok
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01:54:14 **** query probl 13.3.18 chebyshev for z=5 What is the least possible number of elements of a sample which lie within 5 standard deviations of the mean?
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RESPONSE --> 1- 2/5= 0.60
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01:54:28 ** The formula 1 - 1/k^2 gave you .96. That's the proportion which must under any circumstances lie between mean and 5 std dev from the mean. So the number is .96 n, where n is the number of elements in the sample. **
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RESPONSE --> ok
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01:55:06 query probl 13.3.48 mean length of stay 2.7 days, std dev 7.1 days.
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RESPONSE -->
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01:55:12 ** A sketch of a normal distribution will be a normal, or bell-shaped curve with its peak at the mean, dropping to about 60% of peak value at 1 std dev from the mean and to about 14% of peak value at 2 std dev from the mean. For a normal distribution with mean 2.7 and std dev. 7.1 one std dev from the mean occurs at 2.7 + 7.1 = 9.8 and at 2.7 7.1 = -4.4; two std dev from the mean occurs at 16.9 and -11.5. The corresponding normal curve cannot represent length of stay, since length of stay must not be less than zero. As a result we obtain a curve which tails off for large values of x, but whose area is concentrated mostly between 0 and 2.7. This curve is not symmetric like the bell curve but is very skewed, bunched up on one side of the mean 2.7 and more spread out for larger values. **
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RESPONSE --> ok
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01:56:05 **** Describe your sketch of the distribution of lengths of stay.
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RESPONSE --> The bell shape would go from 2.7 days at the lowest and 7.1 days at the highest
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01:56:18 2.7 is in the center with each number within 7.1 of the right or left of 2.7 and each additonal number on the left or right within 7.1 of each other. I see the curve as not being skewed.
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RESPONSE --> ok
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01:57:10 **** Is your distribution skewed? If so why, and if not why do you think it shouldn't be?
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RESPONSE --> yes, because of the deviation of 7.1
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01:57:18 STUDENT RESPONSE AND INSTRUCTOR COMMENT: I said no, but I'm not really sure how to determine this if I draw it myself, becasue I place all the numbers at an equal distance from each other. It just hit me while typing this that it must be skewed to the right because I can't really have negative days stay in the hospital. I think I have confused myself ** You didn't confuse yourself. That's exactly the point. You can't stay fewer than 0 days; since even 1 standard deviation is way below 0 the deviations must be primarily to the right of the mean. So the distribution must be skewed significantly to the right. GENERAL SUGGESTION: In general to understand the graphs of various distributions, try to understand in terms first of the bell-shaped curve with max height at the mean, dropping to about 60% height at a distance of 1 std dev from the mean and to about 14% at 2 std dev from the mean. Then understand that this distribution can be distorted, or skewed, as in this problem. This occurs when most of the distribution lies close to the mean on one side, with a smaller part of the distribution spread out further from the mean on the other. **
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RESPONSE --> ok
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01:57:38 017. `query 17
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RESPONSE --> ok
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