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 14.5 in. and a standard deviation of 0.9 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:

16.59 inches

Step-by-step explanation:

Mean value = u = 14.5 inches

Standard deviation = $\sigma$ = 0.9 in

We need to find the 99th percentile of the given distribution. This can be done by first finding the z value associated with 99th percentile and then using that value to calculate the exact value of hip breadth that lies at 99th percentile

From the z-table, the 99th percentile value is at a z-value of:

z = 2.326

This means 99% of the z-scores are below this value. Now we need to find the equivalent hip breadth for this z-score

The formula to calculate the z score is:

$z=\frac{x-u}{\sigma}$

where, x is the hip breadth which is equivalent to this z-score.

Substituting the values we have:

$2.326=\frac{x-14.5}{0.9}\\\\ 2.0934=x-14.5\\\\ x=16.5934$

Rounded to 2 decimal places, engineers should design the seats which can fit the hip breadth of upto 16.59 inches to accommodate the 99% of all males.