Question

Engineers want to design seats in commercial aircraft so that they are wide enough to fit 95% of all males.

(Accommodating 100% of males would require very wide seats that would be much too expensive.) Men have
hip breadths that are normally distributed with a mean of 14.2 in, and a standard deviation of 0.9 in. Find P95-
That is, find the hip breadth for men that separates the smallest 95% from the largest 5%.
RECEICES
The hip breadth for men that separates the smallest 95% from the largest 5% is P95
(Round to one decimal place as needed.)
in.

Answer 1

To find the hip breadth for men that separates the smallest 95% from the largest 5%, we can use the standard normal distribution table. First, convert the hip breadth to a z-score, then find the z-score that corresponds to the 95th percentile using the standard normal distribution table. Finally, solve for the hip breadth by plugging in the values.

Step-by-step explanation:

To find the hip breadth for men that separates the smallest 95% from the largest 5%, we can use the standard normal distribution table. First, we need to convert the hip breadth to a z-score by subtracting the mean and dividing by the standard deviation. Then we can use the z-score to find the corresponding percentile using the standard normal distribution table. In this case, we want to find the z-score that corresponds to the 95th percentile.

Using the formula for z-score: z = (x - mean) / standard deviation

z = (P95 - 14.2) / 0.9

Next, we can use the standard normal distribution table or a calculator to find the z-score that corresponds to the 95th percentile, which is approximately 1.645. Plugging in the values, we get:

1.645 = (P95 - 14.2) / 0.9

Solving for P95, we get:

P95 = (1.645 * 0.9) + 14.2 = 15.5

Therefore, the hip breadth for men that separates the smallest 95% from the largest 5% is 15.5 inches.