Problem
11) Suppose that an airline uses a seat width of 16.3 in. Assume men have hip breadths that are normally distributed with a mean of \( 14.1 \mathrm{in} \). and a standard deviation of 1in. Complete parts(a) through (c) below. (a) Find the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.3. (Answer: 0.0139) 1 2 (b) If a plane is filled with 123 randomly selected men, find the probability that these men have a mean hip breadth greater than 16.3in. (Answer: 0.0000 ) (c) Which result should be considered for any changes in seat design: the result from part (a) or part (b)? Please explain.
Solution
Step 1 ：\( P(X > 16.3) = 1 - P(X \le 16.3) \)
Step 2 ：\( z = \frac{16.3 - 14.1}{1} = 2.2 \)
Step 3 ：\( P(X > 16.3) = 1 - P(Z \le 2.2) = 0.0139 \)
Step 4 ：\( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt{123}} = 0.0902 \)
Step 5 ：\( z = \frac{16.3 - 14.1}{0.0902} = 24.39 \)
Step 6 ：\( P(\bar{X} > 16.3) = 1 - P(Z \le 24.39) = 0.0000 \)
