course Mth 152 ???????G??????assignment #011
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16:47:50 **** 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 --> 0 points - 6 times 1 points - 4 times 2 points - 7 times .30 for 0 points .21 for 1 points .49 for 2 points Theorectical .3/6 = .05 .21/4 = .05 .49/2 = .24 Empirical 6/.3 = 20 4/.21=19.04 2/.49= 4.08
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16:49:51 ** 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 --> I divided the numbers instead of mutipliying. Also, I did not treat each as a 1 & 2 as 2 shots
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17:04:31 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 --> right 14 times, left 4 times & straight 5 times 1/2*14= 14/2 = 7 1/6*4= 4/6 = 2/3 1/3*5 = 5/3 = 1 2/3
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17:04:46 ** 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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17:05:42 **** 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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17:05:43 **** 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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17:05:46 ** 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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17:05:47 query probl 13.3.48 mean length of stay 2.7 days, std dev 7.1 days.
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17:05:48 ** A sketch of a normal distribution will be a normal, or ?ell-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 ?ails 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, ?unched up?on one side of the mean 2.7 and more spread out for larger values. **
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17:05:50 **** Describe your sketch of the distribution of lengths of stay.
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17:05:52 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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17:05:54 **** Is your distribution skewed? If so why, and if not why do you think it shouldn't be?
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17:05:56 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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17:05:58 017. `query 17
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17:06:00 **** query problem 13.4.12 z score for KG's rebounds (.4 from bottom range 10-13)
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17:06:04 ** The z score for KG is his total number of rebounds minus the mean average number of rebounds for all the players and then divided by the standard deviation. In KG' s case: z = (489 - 538.2) / 38.8 = -1.3 **
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17:06:05 ** The z score for KG is his total number of rebounds minus the mean average number of rebounds for all the players and then divided by the standard deviation. In KG' s case: z = (489 - 538.2) / 38.8 = -1.3 **
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17:06:07 query problem 13.4.30 midquartile same as median? (Q1+Q3)/2
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17:06:08 ** If the median is the actual number in the middle, the it's not necessarily equal to the mean of the first and third quartile. There are different ways to see this. For example suppose that in a large set of numbers, the median number is at least 2 greater than the next smaller number and 2 smaller than the next greater number. Then if all the other numbers stay the same, but the median is increased or decreased by 1, it's still in the middle, so it's still the median. Since all the other numbers stay the same, the first and third quartiles are the same as before, so (Q1 + Q3) / 2 is still the same as before. However the median has changed. So if the median was equal to (Q1 + Q3) / 2, it isn't any more. And if it is now, it wasn't before. In either case we see that the median is not necessarliy equal to the midquartile. To be even more specific, the median of the set {1, 3, 5, 7, 9, 11, 13} is 7. The median of the set {1, 3, 5, 8, 9, 11, 13} is 8. The midquartile of both sets is the same, so for at least one of the two sets (namely the second, as you can verify for yourself) the median and the midquartile are different. **
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17:06:11 018. `query 18
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17:06:12 018. `query 18
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17:06:15 **** query problem 13.5.12 percent above 115 IQ, mean 100 std dev 15
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17:06:16 ** The z-score is measured relative to the mean. The mean is 100, and you need to measure the z score of 115. 115 is 15 units from the mean, which gives you a z-score of 15 / 15 = 1. The table tells you that .339 of the distribution lies between the mean and z = 1. You want the proportion beyond 115. Since half the distribution lies to the right of the mean, and .339 of the distribution lies between the mean and z = 1, we conclude that .5 - .339 = .159 of the distribution lies to the right of z = 1. It follows that .159, or 15.9% of the distribution exceeds an IQ of 115. **
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17:06:18 **** query problem 13.5.20 area between z=-1.74 and z=-1.14
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17:06:20 ** According to the table the z score for -1.74 is .373 and the z score for -1.14 is .459, meaning that .373 of the distribution lies between the mean and z = -1.73 and .459 of the distribution lies between the mean and z = -1.14. Since -1.74 and -1.14 both lie on the same side of the mean, the region between the mean and -1.74 contains the region between the mean and -1.14. The region lying between z = -1.14 and z = -1.74 is therefore that part of the .459 that doesn't include the .373. The proportion between z = -1.14 and z = -1.74 is therefore .459 - .373 = .086, or 8.6%. **
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17:06:21 ** According to the table the z score for -1.74 is .373 and the z score for -1.14 is .459, meaning that .373 of the distribution lies between the mean and z = -1.73 and .459 of the distribution lies between the mean and z = -1.14. Since -1.74 and -1.14 both lie on the same side of the mean, the region between the mean and -1.74 contains the region between the mean and -1.14. The region lying between z = -1.14 and z = -1.74 is therefore that part of the .459 that doesn't include the .373. The proportion between z = -1.14 and z = -1.74 is therefore .459 - .373 = .086, or 8.6%. **
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17:06:24 **** query problem 13.5.30 of 10K bulbs, mean lifetime 600 std dev 50, # between 490 and 720 **** How many bulbs would be expected to last between 490 and 720 hours? **** Describe in detail how you obtained your result.
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17:06:25 ** You first calculate the z value for each of the given lifetimes 490 hrs and 720 hrs. You should then sketch a graph of the distribution so you can see how the regions are located within the distribution. Then interpret what the table tells you about the proportion of the distribution within each region and apply the result to the given situation. The details: The displacement from the mean to 490 is 490 - 600 = - 110 (i.e., 490 lies 110 units to the left of the mean). The z value corresponding to 490 hours is therefore z = -110/50 = -2.2. The area of the region between the mean and z = -2.2 is found from the table to be .486. Similarly 720 lies at displacement 720- 600 = 120 from the mean, giving us z = 120/50 = 2.4. The area of the region between the mean and z = 2.4 is shown by the table to be .492. Since one region is on the negative side and the other on the positive side of the mean, the region lying between z = -2.2 and z = 2.4 contains .486 + .492 = .978 of the distribution. Out of 10,000 bulbs we therefore expect that .978 * 10,000 = 9780 of the bulbs will last between 490 and 720 hours. **
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17:06:26 **** query problem 13.5.48 A's for > mean + 3/2 s What percent of the students receive A's, and how did you obtain your result?
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17:06:30 ** A's are given for z scores greater than 1.5. The area between mean and z = 1.5 is given by the table as .433. To the right of z = 1.5, corresponding to the A's, we have .500 - .433 = .067 or 6.7% of the total area. So we expect that 6.7% of the group will receive A's. **
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17:06:32 GENERAL ADVICE: To solve problems of the type covered in this section it is a good idea to follow a strategy something like the following: 1. Find the z-score(s) corresponding to the given values. 2. Look up the corresponding numbers on the table. 3. Sketch a graph of the normal distribution representing what the numbers in the table tell you. Be sure you understand that the table tells you the proportion of the distribution lying between the mean and the given z value. 4. Decide what region of the graph corresponds to the result you are trying to find. 5. Find the proportion of the total area lying within this region. 6. If necessary apply this proportion to the given numbers to get your final result. See how this procedure is applied in the given solutions. Then you should probably rework the section, being sure your answers agree with those given in the back of the text, and send me questions about anything you aren? sure you understand.
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17:06:34 019. `query 19
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17:06:37 query problem 13.6.9 wt vs ht 62,120; 62,140; 63,130; 65,150; 66,142; 67,13068,175; 68,135; 70,149; 72,168. Give the regression equation and the predicted weight when height is 70.
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17:06:39 ** The equation is obtained by substituting the weights for y and the heights for x in the formula for the regression line. You get y = 3.35 x - 78.4. To predict weight when height is 70 you plug x = 70 into the equation: y = 3.35 * 70 - 78.4. You get y = 156, so the predicted weight for a man 70 in tall is 156 lbs. **
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17:06:42 **** query problem 13.6.12 reading 83,76, 75, 85, 74, 90, 75, 78, 95, 80; IQ 120, 104, 98, 115, 87, 127, 90, 110, 134, 119
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17:06:46 **** query problem 13.6.15 years 0-5, sales 48, 59, 66, 75, 80, 89 What is the coefficient of correlation and how did you obtain it?
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17:06:47 **STUDENT SOLUTION: X Y XY X^2 Y^2 0 48 0 0 2304 1 59 59 1 3481 2 66 132 4 4356 3 75 225 9 5626 4 80 320 16 6400 5 90 450 25 8100 Sums= 15 418 1186 55 30266 The coefficient of the correlation: r = .996 I found the sums of the following: x = 15, y = 418, x*y = 1186, x^2 = 55 n = 6 because there are 6 pairs in the data I also had to find Ey^2 = 30266 I used the following formula: r = 6(1186) - 15(418) / sq.root of 6(55) - (15)^2 * sq. root of 6(30266) - (418)^2 = 846 / sq. root of 105 * 6872 = 846 / sq. root of 721560 = 846 / 849.4 = .996 **
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17:06:52 ** STUDENT SOLUTION: Calculating sums and regression line: n = 8 sum x = 56 sum x^2 = 560 sum = 77.7 sum y^2 = 1110.43 sum xy = 786.4 a = 1.44 b = -.39 r = .99 In the year 2010 the x value will be 16. y' = 1.44(16) - .39 = 22.65. There is an expected 22.65% increase in population by the year 2010. **
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17:06:53 020. `query 20
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