Question

Engineers want to design seats in commercial aircraft so that they are wide enough to fit 90% of all males. (Accommodating 100% of males would require very wide seats that would be much too expensive.) Men have hip breadths that are nomally distributed with a mean of 14.1 in. and a standard deviation of 0.9 in. Find Pao. That is, find the hip breadth for men that separates the smallest 90% from the largest 10%. The hip breadth for men that separates the smallest 90% from the largest 10% is P90= in. %3D (Round to one decimal place as needed.)

Answer 1

The z-score for the 90th percentile is z = 1.282.

Using the formula for z-score, separate x

$\begin{gathered} z=\frac{x-\mu}{\sigma} \\ \sigma z=x-\mu \\ \mu+\sigma z=x \\ x=\mu+\sigma z \end{gathered}$

Given the following

$\begin{gathered} \mu=14.1 \\ \sigma=0.9 \\ z=1.282 \end{gathered}$

Solve for x

$\begin{gathered} x=\mu+\sigma z \\ x=14.1+(0.9)(1.282) \\ x=14.1+2.182 \\ x=16.282 \end{gathered}$

Rounding the answer to one decimal place, the 90th percentile for hip breadth is 16.3 inches.