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 14.6 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 $p_{90}=$ in (round to one decimal place as needed.)

Explanation:

Step1: Find the z - score

We want to find the z - score corresponding to a cumulative probability of 0.90. Looking up in the standard normal distribution table (z - table), the z - score $z$ such that $P(Z

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value we want to find. We know that $\mu = 14.6$ in, $\sigma=0.9$ in, and $z = 1.28$. Rearranging the formula for $x$ gives $x=\mu + z\sigma$.
Substitute the values: $x=14.6+1.28\times0.9$.

Step3: Calculate the value of $x$

First, calculate $1.28\times0.9 = 1.152$. Then, $x=14.6 + 1.152=15.752$.

Answer:

$15.8$ in