Question

Engineers want to design seats in commercial aircraft so that they are wide enough to fit 90 ​% 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 14.5    in. and a standard deviation of 1.1 in.

Find Upper P 90 . That​ is, find the hip breadth for men that separates the smallest 90 ​% from the largest 10 ​%. The hip breadth for men that separates the smallest 90 ​% from the largest 10 ​% is Upper P 90 equals? in inches

Answer 1

Solution:-

Given that,

mean = = 14.5

standard deviation = = 1.1

Using standard normal table,

P(Z > z) = 90%

= 1 - P(Z < z) = 0.90  

= P(Z < z) = 1 - 0.90

= P(Z < z ) = 0.10

= P(Z <-1.28 ) = 0.10  

z =-1.28

Using z-score formula,

x = z * +

x = -1.28* 1.1+14.5

x = 13.092