Let X be the hip breadth for men. X follows Normal distribution with mean μ =14.8 and standard deviation σ =1.1
Here we have to find the hip breadth for men that separates the smallest 90% from the largest 10%
That is we have to find X that separates lower 90% area from upper 10%. For that we will first find z such that probability below it is 0.9 and above it 0.1
P(Z < z) =0.9
Using z score table to find z value
We do not have probability value exactly 0.9 so we will take probability close to 0.9 which is 0.8997 and corresponding z score is 1.28
So we have P(Z < 1.28) =0.9
Now using z=1.28, μ =14.8 and σ =1.1 we will find value of x
x = z*σ + μ
= (1.28 * 1.1) + 14.8
x = 16.208 rounding to nearest integer x= 16
The hip breadth that separates lower 90% area from upper 10% is 16 inch
