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 14.2 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

Answer:

The answer is 15.68 in

Explanation:

X~N(μ=14.2; σ=0.9)

$P(X\leq a)=0.95$

I want to find "a", so:

$P(0\leq X\leq a)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{a-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}$

Z~N(0; 1)

$P(0\leq X\leq a)=P(\frac{0-14.2}{0.9} \leq Z}\leq \frac{a-14.2}{0.9})=\phi(\frac{a-14.2}{0.9})-\phi(-15.78)=0.95$

Φ(-15.78)≅0

$\phi(\frac{a-14.2}{0.9})=0.95$

Based on the Standard Normal table

$\phi(Z_{a} )=0.95$ if $Z_{a}=1.64485$

(a-14.2)/0.9=1.64485, then a=15.68