In Exercises 13–20, use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.)

	Mean	St.Dev.	Distribution
Males	23.5 in	1.1 in	Normal
Females	22.7 in	1.0 in	Normal

For males, find $P_{90}$, which is the length separating the bottom 90% from the top 10%.

The data for sitting back-to-knee length for adult males and females are provided.

The left tailed area is equal to the cumulative probabilities, which are obtained by using the standard normal table (Table A-2) for z-scores.

In the case of finding the right-tailed areas, the difference of these cumulative probabilities from 1 gives the required area towards the right of the z-score.

Let X represent the male back-to-length.

Let x be the value of length separating the bottom 90% of lengths from the top 10%, with a corresponding z-score z.

The shaded area in the graph shows the 90% bottom region corresponding to value x.

Then,

$$\begin{gathered} P\left(X<x\right)=0.90 \\ P\left(Z<z\right)=0.90 \end{gathered}$$

Where

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

Use the standard normal table;the area of 0.90 is observed corresponding to the row value 1.2 and the column value 0.08. This implies that the z score is 1.28.

Mathematically,

$P\left(Z<1.28\right)=0.90$

Thus, the value of z is 1.28.

The length is computed as:

$$\begin{gathered} \frac{x-23.5}{1.1}=1.3 \\ x=\left(1.3\times 1.1\right)+23.5 \\ =24.93 \\ \approx 24.9 \end{gathered}$$

Therefore, the length separating the bottom 90% from the top 10% is $P_{90}=24.9\text{in}$.