Question

engineers want to design seats in commercial aircraft so that they are wide enough to fit 99% of all adults. (accommodating 100% of adults would require very wide seats that would be much too expensive.) assume adults have hip widths that are normally distributed with a mean of 14.6 in. and a standard deviation of 0.9 in. find p99. that is, find the hip width for adults that separates the smallest 99% from the largest 1%. what is the maximum hip width that is required to satisfy the requirement of fitting 99% of adults? in. (round to one decimal place as needed.)

Explanation:

Step1: Find the z - score

We want to find \(P_{99}\), which means the area to the left of the z - score is \(0.99\). Using a standard normal table or a calculator with a normal - distribution function (such as the invNorm function on a TI - 84 Plus: invNorm\((0.99,0,1)\)), the z - score \(z\) corresponding to an area of \(0.99\) is approximately \(z = 2.33\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 14.6\) (mean), \(\sigma=0.9\) (standard deviation), and \(z = 2.33\).
We need to solve for \(x\). Rearranging the formula gives \(x=\mu+z\sigma\).
Substitute the values: \(x = 14.6+(2.33\times0.9)\).
First, calculate \(2.33\times0.9=2.097\).
Then, \(x=14.6 + 2.097=16.697\).

Answer:

\(16.7\) in.