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

Find the probability that a female has a back-to-knee length greater than 24.0 in.

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

Let Y represent the female back-to-knee length.

Thus,

$$\begin{gathered} Y~N\left(\mu ,{\sigma }^2\right) \\ ~N\left(22.7,{1.0}^2\right) \end{gathered}$$

The back-to-knee length of females is 24 in.

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.

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

4. Shade the region right of 24.

The area under the curve has a one-to-one correspondence with the probabilities.

The shaded region represents the probability that the back-to-knee length is greater than 24.0 in.

The z score is computed as:

$$\begin{gathered} z=\frac{y-\mu }{\sigma } \\ =\frac{24-22.7}{1} \\ =1.3 \end{gathered}$$

Therefore, the z score is 1.30.

By using the standard normal table, the area to the left of 1.30 is obtained from the table in the intersection cell with row value 1 and the column value 0.30, which is obtained as 0.9032.

The probability that a female has a back-to-knee length greater than 24 in. is computed as:

$$\begin{gathered} P\left(X>24.0\right)=P\left(z>1.30\right) \\ =1-P\left(z<1.30\right) \\ =1-0.9032 \\ =0.0968 \end{gathered}$$

Thus, the probability that a female has a back-to-knee length greater than 24 in. is 0.0968.