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 14.2 in. and a standard deviation of 0.90.9 in. Find Upper P99. 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 1% is P99= ___in.

Answer 1

Solution :

Given that,

mean = = 14.2

standard deviation = = 0.90.9

Using standard normal table,

P( Z < z) = 699%
P(Z < z) = 0.99

z = 2.33

Using z-score formula,

x = z * +

x = 2.33 * 0.909 + 14.2

=16.317

P₉₉ = 16.32