Question

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 14.1 in. and a standard deviation of 1.1 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 (p_{99}=square) in.
(round to one decimal place as needed.)

Explanation:

Step1: Use z - score formula

We use the standard normal distribution $Z=\frac{X-\mu}{\sigma}$, where $X$ is the value from the original normal distribution, $\mu$ is the mean, $\sigma$ is the standard deviation, and $Z$ is the z - score. We want to find the z - score $z$ corresponding to a cumulative probability of $0.99$. Looking up in the standard normal distribution table (or using a calculator with a normal - distribution function), the z - score $z$ such that $P(Z\leq z)=0.99$ is approximately $z = 2.33$.

Step2: Rearrange the z - score formula to solve for $X$

Starting with $Z=\frac{X-\mu}{\sigma}$, we can solve for $X$: $X=\mu + Z\sigma$. We know that $\mu = 14.1$ inches and $\sigma=1.1$ inches, and $Z = 2.33$.
Substitute the values into the formula: $X=14.1+2.33\times1.1$.

Step3: Calculate the value of $X$

First, calculate $2.33\times1.1 = 2.563$. Then, $X=14.1 + 2.563=16.663$.

Step4: Round the result

Rounding $16.663$ to one decimal place, we get $X\approx16.7$.

Answer:

$16.7$