: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, the 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

Find the probability that a male has a back-to-knee length less than 21 in.

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

Let X represent the male back-to-knee-length.

Then,

$$\begin{gathered} X~N\left(\mu ,{\sigma }^2\right) \\ ~N\left(23.5,{1.1}^2\right) \end{gathered}$$

Steps to make a normal curve are as follows:

1. Make a horizontal and vertical axis.

2. Mark the points 20, 22, 24, and 26 on the horizontal axis and points 0.1, 0.2, 0.3, and 0.4.

3. Provide titles to the horizontal and vertical axis as “x” and “f(x),” respectively.

4. Shade the region lesser than 21.

Due to the area of 1 under the standard normal curve, there is a one-to-one correspondence between area and probability.The shaded area represents the probability that back-to-knee length is 21 in.

The z score is computed as:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{21-23.5}{1.1} \\ =-2.273 \\ \approx -2.27 \end{gathered}$$

Therefore, the z score is -2.27.

By using the standard normal table, the area to the left of -2.27is obtained from the table in the intersection cell with row value -2 and the column value 0.27, which is obtained as 0.0116.

The probability that a male has a back-to-knee length less than 21 in. is computed as:

$P\left(z<-2.27\right)=0.0116$

Thus, the probability that a male has a back-to-knee length less than 21 in. is 0.0116.