Question

engineers want to design seats in commercial aircraft so that they are wide enough to fit 95% 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 p95. that is, find the hip breadth for men that separates the smallest 95% from the largest 5%.

the hip breadth for men that separates the smallest 95% from the largest 5% is p95 = □ in. (round to one decimal place as needed.)

Explanation:

Step1: Find the z - score

We want to find the z - score corresponding to the 95th percentile. 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 $x$ is the value from the original normal distribution, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know $\mu = 14.7$, $\sigma=0.9$ and $z = 1.645$. Rearranging the formula for $x$ gives $x=\mu + z\sigma$.

Step3: Calculate the value of $x$

Substitute the values into the formula: $x=14.7+1.645\times0.9$.
$x = 14.7+1.4805=16.1805$.

Answer:

$16.2$