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 14.2 in. and a standard deviation of 1.2 in. Find Upper P 99. That​ is, find the hip breadth for men that separates the smallest 99​% from the largest 1​%.

Answer 1

Answer:

Upper P99 = 17 in.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 14.2, \sigma = 1.2$

Upper P99

This is X when Z has a pvalue of 0.99. So X when Z = 2.327.

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

$2.327 = \frac{X - 14.2}{1.2}$

$X - 14.2 = 1.2*2.327$

$X = 17$

So

Upper P99 = 17 in.