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 14.3 in. And a standard deviation of 0.9 in. Find p99. That​ is, find the hip breadth for men that separates the smallest ​99% from the largest ​1%. The hip breadth for men that separates the smallest ​99% from the largest ​1% is p99= how many inches

Answer 1

Answer:

p99 = 16.4 inches

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 14.3 in. And a standard deviation of 0.9 in.

This means that $\mu = 14.3, \sigma = 0.9$

Find p99.

This is the value of X when Z has a p-value of 0.99, so X when Z = 2.327. Then

$Z = \frac{X - \mu}{\sigma}$

$2.327 = \frac{X - 14.3}{0.9}$

$X - 14.3 = 0.9*2.327$

$X = 16.4$

So

p99 = 16.4 inches