Question

Engineers want to design seats in commercial aircraft so that they are wide enough to fit 99​% 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 14.3 in. and a standard deviation of 0.8 in. Find Upper P99. That​ is, find the hip breadth for men that separates the smallest 99​% from the largest 1​%.

Answer 1

Solution:-

Given that,

mean = = 14.3

standard deviation =  = 0.8

Using standard normal table,

P(Z > z) = 99%

= 1 - P(Z < z) = 0.99

= P(Z < z) = 1 - 0.99

= P(Z < z ) = 0.01

= P(Z < -2.33 ) = 0.01

z =-2.33

Using z-score formula,

x = z * +

x = -2.33 * 0.8+14.3

x = 12.436