Suppose that an airline uses a seat width of 16.6 in. Assume men have hip breadths that are normally distributed with a mean of 15 in. and a standard deviation of 0.9 in. 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 16.6 in. The probability is (Round to four decimal places as needed.) (b) If a plane is filled with 114 randomly selected men, find the probability that these men have a mean hip breadth greater than 16.6 in. The probability is (Round to four decimal places as needed.) (c) Which result should be considered for any changes in seat design: the result from part (a) or part (b)? The result from should be considered because

Step 1 ：Given that the mean hip breadth of men is 15 in. with a standard deviation of 0.9 in., we are asked to find the probability that a randomly selected man has a hip breadth greater than the seat width of 16.6 in. This involves calculating the Z-score for a hip breadth of 16.6 in. and finding the area under the normal distribution curve to the right of this Z-score.

Step 2 ：The Z-score is calculated as \((seat\_width - mean\_hip\_breadth) / std\_dev\), which gives a Z-score of approximately 1.78.

Step 3 ：Using a standard normal distribution table or a Z-score calculator, we find that the probability corresponding to a Z-score of 1.78 is approximately 0.0377. Therefore, the probability that a randomly selected man has a hip breadth greater than 16.6 in. is approximately \(\boxed{0.0377}\).

Step 4 ：We are also asked to find the probability that the mean hip breadth of 114 randomly selected men is greater than 16.6 in. This involves calculating the standard error, which is the standard deviation divided by the square root of the sample size, and then finding the Z-score for a mean hip breadth of 16.6 in. using this standard error.

Step 5 ：The standard error is calculated as \(std\_dev / \sqrt{sample\_size}\), which gives a standard error of approximately 0.0843.

Step 6 ：The Z-score is then calculated as \((seat\_width - mean\_hip\_breadth) / std\_error\), which gives a Z-score of approximately 18.98.

Step 7 ：Using a standard normal distribution table or a Z-score calculator, we find that the probability corresponding to a Z-score of 18.98 is approximately 0.0. Therefore, the probability that the mean hip breadth of 114 randomly selected men is greater than 16.6 in. is approximately \(\boxed{0.0}\).

Step 8 ：For any changes in seat design, the result from part (a) should be considered because it represents the probability of a single man having a hip breadth greater than the seat width, which is a more direct measure of whether the seat width is adequate.