Engineers want to design seats in commercial aircraft so that they are wide enough to fit 9090% 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 14.814.8 in. and a standard deviation of 1.11.1 in. find upper p 90p90. that is, find the hip breadth for men that separates the smallest 9090% from the largest 1010%. the hip breadth for men that separates the smallest 9090% from the largest 1010% is upper p 90p90equals= nothing in.

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

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