Question

Engineers want to design seats in commercial aircraft so that they are wide enough to fit 9999​% 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 14.514.5  in. and a standard deviation of 0.80.8 in. Find Upper P 99P99. That​ is, find the hip breadth for men that separates the smallest 99​% from the largest 11​%.

The hip breadth for men that separates the smallest 99% from the largest 11​% is Upper P 99P99equals=nothing in. ​(Round to one decimal place as​ needed.)

Answer 1

Given that,

mean = = 14.5

standard deviation = = 0.8

Using standard normal table,

P(Z < z) = 99%

= P(Z < z) = 0.99

= P(Z <2.33 ) = 0.99

z = 2.33

Using z-score formula  

x= z * +

x= 2.33 *0.8+14.5

x= 16.4