STA 301 - HW07 Key - STA 301 Applied Statistics Spring 2008...

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-3 -2 -1 0 1 2 3 .8198 .0901 .0901 STA 301 – Applied Statistics Spring 2008, Section B Homework #7 - Key Due Wednesday, March 12 Question 1 : Suppose Z is a random variable having a standard normal distribution. Calculate the following probabilities: (a) ( 65 . 1 Z P (b) ( 65 . 1 0 < < Z P 1 – P(Z < 1.65) = 1 - .9505 = .0495 P(0 < Z < 1.65) = P(Z > 0) – P(Z > 1.65) = .5000 - .0495 = .4505 (c) ( 34 . 1 34 . 1 < - Z P (d) ( 75 . 1 25 . 1 < < - Z P P(-1.34 < Z < 1.34) = 1 – 2*P(Z < -1.34) P(-1.25 < Z < 1.75) = 1 – 2 * .0901 = .8198 = P(Z < 1.75) – P(Z < -1.25) = .9599 - .1056 = .8543 -3 -2 -1 0 1 2 3 .0495 .9505 -3 -2 -1 0 1 2 3 .5000 .4505 .0495 -3 -2 -1 0 1 2 3 .1056 .0401 .8543
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STA 301-B, Spring 2008 Homework 7 Key Page 2 of 5 -3 -2 -1 0 1 2 3 .0594 .9406 -3 -2 -1 0 1 2 3 .7301 .1587 .1112 (e) ( 1 | 5 . 1 < Z Z P P(Z < 1.5 | Z > 1) = P(Z < 1.5 and Z > 1) / P(Z > 1) ** Definition of conditional probability = P(1 < Z < 1.5) / P(Z > 1) ** Solid area divided by total shaded area Question 2 As reported in Runner’s World magazine, the times of the finishers in the New York City 10km run are normally distributed with a mean of 61 minutes and a standard deviation of 9 minutes. Let X be the time (in minutes) of a randomly selected finisher. Find: (a) ( 75 X P (b) ( 70 50 < X X P U ( ( ( 29 5791 .
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