Question

Engineers want to design seats in commercial aircraft so that they are wide enough to fit 98% 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 14.4 in. and a standard deviation of 1.0 in.. Find P98. That is, find the hip breadth for men that separates the smallest 98% from the largest 2

Answer 1

P98 = 16.154in

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Men have hip breadths that are normally distributed with a mean of 14.4 in. and a standard deviation of 1.0 in.

This means that
`\mu = 14.4, \sigma = 1`

Find P98

This is the 98th percentile, that is, X when Z has a pvalue of 0.98, so X when Z = 2.054.

`Z = (X - \mu)/(\sigma)`

`2.054 = (X - 14.1)/(1)`

`X = 2.054 + 14.1`

`X = 16.154`

So

P98 = 16.154in