Engineers want to design seats in commercial aircraft so that they are wide enough to fit $99 \%$ 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 14.7 in. and a standard deviation of 0.9 in. Find $P_{99}$. That is, find the hip breadth for men that separates the smallest $99 \%$ from the largest $1 \%$.
The hip breadth for men that separates the smallest $99 \%$ from the largest $1 \%$ is $\mathrm{P}_{99}=$ in. (Round to one decimal place as needed.)

Step 1 ：The problem is asking for the 99th percentile of the normal distribution. This is a standard problem in statistics and can be solved using the z-score formula. The z-score is the number of standard deviations a particular value is from the mean. In this case, we want to find the value that is at the 99th percentile, which means it is higher than 99% of all other values.

Step 2 ：We are given that the mean hip breadth is 14.7 inches and the standard deviation is 0.9 inches. We also know that the z-score for the 99th percentile is approximately 2.326.

Step 3 ：We can use the z-score formula to find the value that corresponds to the 99th percentile. The formula is \(Z = \frac{X - \mu}{\sigma}\), where \(Z\) is the z-score, \(X\) is the value we are trying to find, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 4 ：Rearranging the formula to solve for \(X\), we get \(X = Z\sigma + \mu\). Substituting the given values, we get \(X = 2.326 * 0.9 + 14.7\).

Step 5 ：Calculating the above expression, we find that \(X = 16.8\).

Step 6 ：So, the hip breadth for men that separates the smallest 99% from the largest 1% is \(\boxed{16.8}\) inches.