91影视

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

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

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 鈥渪鈥� and 鈥渇(x),鈥� respectively.

4. Shade the region in between 22 and 24.

The area between 22.0 and 24.0 is equal to the probability of getting the length between 22.0 and 24.0.

The probability has a one-to-one correspondence with the area of the curve represented as $P\left(22.0<X<24.0\right)$.

The z-score corresponding to$x_1=22$

$$\begin{gathered} z_1=\frac{x_1-渭}{蟽} \\ =\frac{22-23.5}{1.1} \\ =-1.36 \end{gathered}$$

Therefore, the z score is -1.36.

The z-score corresponding to $y_1=24$,

$$\begin{gathered} z_2=\frac{x_2-渭}{蟽} \\ =\frac{24-23.5}{1.1} \\ =0.45 \end{gathered}$$

Therefore, the z score is 0.45.

The area between the two z-scores is -1.36 and 0.45.

Mathematically,

$$\begin{gathered} \text{Area}\text{between}-1.36\text{and}0.45=\text{Area}\text{to}\text{the}\text{left}\text{of}0.45-\text{Area}\text{to}\text{the}\text{left}\text{of}-1.36 \\ =P\left(Z<0.45\right)-P\left(Z<-1.36\right)...\left(1\right) \end{gathered}$$

Use the standard normal table,

• the area to the left of 0.45 is obtained from the table in the intersection cell with row value 0.4 and the column value 0.05, which is obtained as 0.6736.
• the area to the left of -1.36 is obtained from the table in the intersection cell with row value -1.3 and the column value 0.06, which is obtained as 0.0869.

Mathematically it is expressed as:

$$\begin{gathered} \text{Area}\text{to}\text{the}\text{left}\text{of}0.45=P\left(Z<0.45\right) \\ =0.6736 \\ \text{Area}\text{to}\text{the}\text{left}\text{of}-1.36=P\left(Z<-1.36\right) \\ =0.0869 \end{gathered}$$

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

$$\begin{gathered} \text{Area}\text{between}-1.36\text{and}0.45=0.6736-0.0869 \\ =0.5867 \end{gathered}$$

Therefore, the probability that a male has a back-to-knee length between 22.0 in. and 24.0 in. is0.5867.