assignment #009 009. Binomial Probabilities Liberal Arts Mathematics II 03-18-2009 k xǧUØ assignment #009 009. Binomial Probabilities Liberal Arts Mathematics II 03-18-2009
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12:54:19 `q001. Note that there are 5 questions in this assignment. What is the probability that on two rolls of a fair die, we obtain exactly two 3's?
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RESPONSE --> There are a total of 42 outcomes, therefore the probablity is 2/42 =1/21 confidence assessment: 2
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12:55:32 The probability of obtaining a 3 on a single role is 1/6 (one of the six possible outcomes is a 3). Since the two rolls are independent, it follows that if two dice are rolled the probability of obtaining two 3's is 1/6 * 1/6 = 1/36.
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RESPONSE --> OK I listed all the possibilities.. so I did it the hard way and STILL got it wrong. self critique assessment: 3
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12:56:01 `q002. What is the probability that on three rolls of a fair die, we obtain exactly two 5's?
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RESPONSE --> 1/36 confidence assessment: 2
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12:58:23 On three rolls of a fair die, the two 5's can occur on the first and second, the first and third or the second and third rolls. That is, of the three available positions the two positions in which the 5's occur can occur in C(3,2) = 3 different ways. Since the probability of a 5 on any roll is 1/6 and the probability of not getting a 5 on a roll is 5/6. Any one of the three ways of getting two 5's and one non-5 is therefore (1/6) * (1/6) * (5/6 ) = 5/216. Since each of the three ways to get the desired outcome occurs with probability 5/216, it follows that Probability of exactly two 3's on three rolls = 3 * 5/216 = 15/216 = 5/72.
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RESPONSE --> OK, I didn't multiply the 5/6 in with the other 1/6's, I just assumed that since it only had to be 2 consectutive 5s out of 3 rolls that it would be the same as the problem before, but I understand that you still have to take the 5/6 in consideration, you can't just leave it off. self critique assessment: 2
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13:00:51 `q003. What is the probability that on six rolls of a fair die, we obtain exactly two 5's?
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RESPONSE --> (1/6)(1/6)(5/6)(5/6)(5/6)(5/6)=.01339 confidence assessment: 2
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13:02:26 In order to get exactly two 5's on six rolls of the fair die, we must get two 5's and four results that are not 5. The probability of getting a 5 on any roll is 1/6, and the probability of getting a result other than 5 is 5/6. Therefore given any two positions out of the six the probability of obtaining 5's in two of the positions and non-5's in the remaining four positions is by the Fundamental Counting Principle Probability of 5's in exactly two of the six positions = (1/6) * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) = (1/6)^2 * ( 5/6)^4. There are C(6,2) ways in which the positions of the two 5's can be selected from the six available positions. Thus we have Probability of exactly two 5's on six flips = C(5,2) * (1/6)^2 * (5/6)^4.
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RESPONSE --> self critique assessment: 3
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13:04:00 `q004. If we let p stand for the probability of getting a 5 on a roll of a die and q for the probability of not getting a 5 on a roll, then how would we expressed a probability of getting exactly r 5's on n rolls?
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RESPONSE --> confidence assessment: 0
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13:04:44 By analogy with the preceding problem, we see that to get r 5's on n rolls we must get 5 the total of r times and non-5 a total of (n-r) times. Since probability of getting a 5 is p, the probability of getting 5 a total of r times is represented by p^r. Since the probability of getting a non-5 is q, then the probability of getting a non-5 a total of (n-r) times is represented by q^(n-r). There are C(n, r) ways to place fives in r of n positions, so the probability of getting 5 fives and n non-fives is C(n, r) * p^r * q^(n-r).
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RESPONSE --> self critique assessment: 0
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f^yxVkpTҕ assignment #014 014. mean vs median Liberal Arts Mathematics II 03-18-2009
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13:06:48 `q001. Note that there are 8 questions in this assignment. {}{} What is the average, or mean value, of the numbers 5, 7, 9, 9, 10, 12, 13, and 15? On the average how 'far' is each number from this mean value?
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RESPONSE --> average-10 5,3,2 confidence assessment: 3
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13:07:16 To get the mean value of the numbers, we first note that there are eight numbers. Then we had the numbers and divide by eight. We obtain 5 + 7 + 9 + 9 + 10 + 12 + 13 + 15 = 80. Dividing by 8 we obtain mean = 80 / 8 = 10. The difference between 5 and the mean 10 is 5; the difference between 7 and the mean 10 is 3; the difference between 9 and 10 is 1; the differences between 12, 13 and 15 and the mean 10 are 2, 3 and 5. So we have differences 5, 3, 1, 1, 0, 2, 3 and 5 between the mean and the numbers in the list. The average difference between the mean and the numbers in the list is therefore ave difference = ( 5 + 3 + 1 + 1 + 0 + 2 + 3 + 5 ) / 8 = 20 / 8 = 2.5.
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RESPONSE --> I just listed the differences, I didn't do the average difference self critique assessment: 2
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13:08:42 `q002 What is the middle number among the numbers 13, 12, 5, 7, 9, 15, 9, 10, 8?
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RESPONSE --> 9 is the middle number.. first I put the numbers in order then took the middle number confidence assessment: 3
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13:08:57 It is easier to answer this question if we place the numbers in ascending order. Listed in ascending order the numbers are 5, 7, 8, 9, 9, 10, 12, 13, and 15. We see that there are 9 numbers in the list. If we remove the first 4 and the last 4 we are left with the middle number. So we remove the numbers 5, 7, 8, 9 and the numbers 10, 12, 13, and 15, which leaves the second '9' as the middle number.
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RESPONSE --> self critique assessment: 3
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13:09:09 `q003. On a list of 9 numbers, which number will be the one in the middle? Note that the middle number is called the 'median'.
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RESPONSE --> 9 confidence assessment: 3
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13:10:02 If the 9 numbers are put in order, then we can find the middle number by throwing out the first four and the last four numbers on the list. We are left with the fifth number on the list. In general if we have an odd number n of number in an ordered list, we throw out the first (n-1) / 2 and the last (n-1) / 2 numbers, leaving us with the middle number, which is number (n-1)/2 + 1 on the list. So for example if we had 179 numbers on the list, we would throw out the first (179 - 1) / 2 = 178/2 = 89 numbers on the list and the last 89 numbers on the list, leaving us with the 90th number on the list. Note that 90 = (179 - 1) / 2 + 1, illustrating y the middle number in number (n-1)/2 + 1 on the list.
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RESPONSE --> OK self critique assessment: 3
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13:10:41 `q004. What is the median (the middle number) among the numbers 5, 7, 9, 9, 10, 12, 13, and 15?
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RESPONSE --> 10 confidence assessment: 3
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13:11:04 There are 8 numbers on this list. If we remove the smallest then the largest our list becomes 7, 9, 9, 10, 12, 13. If we remove the smallest and the largest from this list we obtain 9, 9, 10, 12. Removing the smallest and the largest from this list we are left with 9 and 10. We are left with two numbers in the middle; we don't have a single 'middle number'. So we do the next-most-sensible thing and average the two numbers to get 9.5. We say that 9.5 is the middle, or median, number.
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RESPONSE --> OK self critique assessment: 3
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13:12:54 `q005. We saw that for the numbers 5, 7, 9, 9, 10, 12, 13, and 15, on the average each number is 2.5 units from the average. Are the numbers in the list 48, 48, 49, 50, 51, 53, 54, 55 closer or further that this, on the average, from their mean?
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RESPONSE --> 2.2 confidence assessment: 3
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13:13:34 The mean of the numbers 48, 48, 49, 50, 51, 53, 54, and 55 is (48 + 48 + 49 + 50 + 51 + 53 + 54 + 55) / 8 = 408 / 8 = 51. 48 is 3 units away from the mean 51, 49 is 2 units away from the mean 51, 50 is 1 unit away from the mean 51, and the remaining numbers are 2, 3 and 4 units away from the mean of 51. So on the average the distance of the numbers from the mean is (3 + 3 + 2 + 1 + 0 + 2 + 3 + 4) / 8 = 18 / 8 = 2.25. This list of numbers is a bit closer, on the average, then the first list.
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RESPONSE --> I miscounted and got 16/7 instead of 18/8 butI see where I messed up. self critique assessment: 0
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13:15:22 `q006. On a 1-10 rating of a movie, one group gave the ratings 1, 8, 8, 9, 9, 10 while another gave the ratings 7, 7, 8, 8, 9, 10. Find the mean (average) and the median (middle value) of each group's ratings. Which group would you say liked the movie better?
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RESPONSE --> Group 2 liked it better.. Group 1 average was 7.5 and group 2s average was 8.1 confidence assessment: 3
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13:15:31 The mean of the first list is (1 + 8 + 8 + 9 + 9 + 10) / 6 = 45 / 6 = 7.5. The median is obtained a throwing out the first 2 numbers on the list and the last 2 numbers. This leaves the middle two, which are 8 and 9. The median is therefore 8.5. The mean of the numbers on the second list is (7 + 7 + 8 + 8 + 9 + 10) / 6 = 49 / 6 = 8 .16. The median of this list is found by removing the first 210 the last 2 numbers on the list, leaving the middle two numbers 8 and 8. The median is therefore 8. The first group had the higher median and the lower mean, while the second group had the lower median but the higher mean. Since everyone except one person in the first group scored the movie as 8 or higher, and since everyone in both groups except this one individual scored the movie 7 or higher, it might be reasonable to think that the one anomalous score of 1 is likely the result of something besides the quality of the movie. We might also note that this score is much further from the mean that any of the other scores, giving it significantly more effect on the mean than any other score. We might therefore choose to use the median, which limits the otherwise excessive weight given to this unusually low score when we calculate the mean. In this case we would say that the first group liked the movie better.
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RESPONSE --> self critique assessment: 3
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13:17:44 `q007. Suppose that in a certain office that ten employees make $700 per pay period, while five make $800 per pay period and the other two make $1000 per pay period. What is the mean pay per period in the office? What is the median?
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RESPONSE --> mean is $13,000 the median is 9000 confidence assessment: 3
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13:17:56 There are a total of 10 + 5 + 2 = 17 employees in the office. The total pay per pay period is 10 * $700 + 5 * $800 + 2 * $1000 = $13,000. The mean pay per period is therefore $13,000 / 17 = $823 approx.. The median pay is obtained by 'throwing out' the lowest 8 and the highest 8 in an ordered list, leaving the ninth salary. Since 10 people make $700 per period, this leaves $700 as the median.
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RESPONSE --> OK self critique assessment: 3
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13:19:02 `q008. In the preceding problem ten employees make $700 per pay period, while five make $800 per pay period and the other two make $1000 per pay period; we just found that the mean pay per period was $823. On the average, how much to the individual salaries differ from the mean?
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RESPONSE --> 6000,9000,11000 confidence assessment: 3
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13:19:08 The mean was found in the preceding problem to be $823. The deviation of $700 from the mean is therefore $123, the deviation of $800 from the mean is $23 and the deviation of $1000 from the mean is $177. Since $700 is paid to 10 employees, $800 to five and $1000 to two, the total deviation is 10 *$123 + 5 * $23 + 2 * $177 = $1630. The mean deviation is therefore $1630 / 17 = $96, approx..
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RESPONSE --> self critique assessment: 3
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