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

14.8

in. and a standard deviation of

1.1

in. Find

Upper P90.

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 P90 equals = ? in.

Answer 1

SOLUTION:

Given that,

mean = = 14.8

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 ) = 0.1

= P(Z < z ) = 0.1

z = -1.28 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x= -1.28*1.1+14.8

x=13.392

Upper P90 equals = 13.392 in.