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 14.1 in. and a standard deviation of 1.2 in. find p95. that is, find the hip breadth for men that separates the smallest 95% from the largest 5%.
the hip breadth for men that separates the smallest 95% from the largest 5% is p95 = □ in.
(round to one decimal place as needed.)

Explanation:

Step1: Find the z - score

For a 95% confidence level (separating the smallest 95% from the largest 5%), the z - score \(z\) corresponding to the cumulative probability of 0.95. Using a standard normal distribution table or calculator, \(z = 1.645\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 14.1\) (mean), \(\sigma=1.2\) (standard deviation), and we want to find \(x\) (the value of \(P_{95}\)).
Rearrange the formula to \(x=\mu + z\sigma\).
Substitute the values: \(x = 14.1+1.645\times1.2\).
First, calculate \(1.645\times1.2 = 1.974\).
Then, \(x=14.1 + 1.974=16.074\).

Answer:

\(16.1\)