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 normally distributed with a mean of 14.2 in. and a standard deviation of 0.9 in. find ( p_{90} ). 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 ( p_{90}=square ) in.
(round to one decimal place as needed.)

Explanation:

Step1: Find the z - score

We want to find the \(z\) - score corresponding to an area of \(0.9\) (since we want the value that separates the smallest \(90\%\) from the largest \(10\%\)). Using a standard normal table (or a calculator with a normal - distribution function, such as the inverse - norm function on a TI - 84: invNorm(0.9,0,1)), the \(z\) - score \(z\) is approximately \(z = 1.28\).

Step2: Use the z - score formula

The \(z\) - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the value we want to find, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. We know that \(\mu = 14.2\), \(\sigma=0.9\), and \(z = 1.28\).

Rearrange the formula for \(x\): \(x=\mu+z\sigma\)

Substitute the values: \(x = 14.2+1.28\times0.9\)

First, calculate \(1.28\times0.9=1.152\)

Then, \(x = 14.2 + 1.152=15.352\)

Answer:

\(15.4\)