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 14.9  in. and a standard deviation of 0.9 in. Find Upper P 95. That​ is, find the hip breadth for men that separates the smallest 95​% from the largest 5​%.

Answer 1

Given that, mean (μ) = 14.9 in and

standard deviation = 0.9 in

Let X ~ N(14.9, 0.9)

We want to find, the value of x such that, P(X < x) = 0.95

First we find, the z-score such that P(Z < z) = 0.95

Using standard normal z-table we get z-score corresponding probability of 0.95 is z = 1.645

For z = 1.645 we get the value of x as follows :

Therefore, we get, P95 = 16.3805 in

Note : if you want answer rounded to two decimal places, then

P95 = 16.38