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 between 22.0 in. and 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-length.

Thus,

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

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 axes as “y” and “f(y),” respectively.

4. Shade the region in between 22 and 24.

The area between 22 and 24 is equal to the probability of getting the length between 22 and 24.

The probability has a one-to-one correspondence with the area of the curve represented as $P\left(22<Y<24\right)$.

The z score corresponding to $y_1=22$is computed as:

$$\begin{gathered} z_1=\frac{y_1-\mu }{\sigma } \\ =\frac{22-22.7}{1} \\ =-0.7 \end{gathered}$$

Therefore, the z score is -0.7.

The z-score corresponding to $y_2=24$,

$$\begin{gathered} z_2=\frac{y_2-\mu }{\sigma } \\ =\frac{24-22.7}{1} \\ =1.3 \end{gathered}$$

Therefore, the z score is 1.3.

The area between the two z-scores is-0.7 and 1.3.

Mathematically,

$$\begin{gathered} P\left(22<Y<24\right)=\text{Area}\text{between}-0.7\text{and}1.3 \\ =\text{Area}\text{to}\text{the}\text{left}\text{of}1.3-\text{Area}\text{to}\text{the}\text{left}\text{of}-0.7 \\ =P\left(Z<1.3\right)-P\left(Z<-0.7\right)...\left(1\right) \end{gathered}$$

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 1.3 and the column value .00, which is obtained as 0.9032.
• the area to the left of 0.7 is obtained from the table in the intersection cell with row value 0.7 and the column value 0.00, which is obtained as 0.7580.

Mathematically it is expressed as:

$$\begin{gathered} \text{Area}\text{to}\text{the}\text{left}\text{of}1.3=P\left(Z<1.3\right) \\ =0.9032 \\ \text{Area}\text{to}\text{the}\text{left}\text{of}-0.7=P\left(Z<-0.7\right) \\ =1-P\left(Z<0.7\right) \\ =1-0.758 \\ =0.242 \end{gathered}$$

Substitute the values in equation (1);the probability that a female has a back-to-knee length between 22.0 in. and 24.0 in. is given as:

$$\begin{gathered} \text{Area}\text{between}-0.7\text{and}1.3=0.9032-0.242 \\ =0.6612 \end{gathered}$$

Therefore, the probability that a female has a back-to-knee length between 22.0 in. and 24.0 in is 0.6612.