Question

Engineers want to design passenger seats in commercial aircraft so that they are wide enough to fit 95 percent of adult men. Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. Find the 95th percentile of the hip breadth of adult men. Round your answer to one decimal place; add a trailing zero as needed. The 95th percentile of the hip breadth of adult men is [HipBreadth] inches.

Answer 1

Given that,

mean = = 14.4

standard deviation = =1.1

Using standard normal table,

P(Z < z) = 95%

= P(Z < z) = 0.95  

= P(Z < 1.64) = 0.95

z = 1.64   

Using z-score formula  

x= z * +

x= 1.64*1.1+14.4

x= 16.204

x=16