In Exercises 5–8, find the area of the shaded region. The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler IQ test).

A shaded region is shown in the graph for the standard normal distribution of the IQ scores of adults.

The mean IQ score is 100.

The standard deviation of the IQ score is 15.

The left-tailed area is equal to the cumulative probabilities that 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 toward the right of the z score.

Let X represent the IQ score of adults.

The variable x is normally distributed with the mean $\mu =100$, and the standard deviation is $\sigma =15$.

The IQ scores marked on the graph are $x=79$and $x=133$.

The z score is computed as shown below.

For $x=79$,

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{79-100}{15} \\ =-1.4 \end{gathered}$$

Therefore, the z score is -1.4.

For $x=133$,

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{133-100}{15} \\ =2.2 \end{gathered}$$

Therefore, the z score is 2.2.

It is required to compute the area between 79 and 133. It is corresponding to the area between two z-scores: -1.2 and 2.2.

Mathematically,

$$\begin{gathered} \text{Area}\text{between}-1.2\text{and}2.2=\text{Area}\text{to}\text{the}\text{left}\text{of}2.2-\text{Area}\text{to}\text{the}\text{left}\text{of}-1.4 \\ =P\left(Z<2.2\right)-P\left(Z<-1.4\right)...\left(1\right) \end{gathered}$$

Using the standard normal table,

• the area to the left of 2.2 is obtained from the table in the intersection cell with row value 2 and column value 0.2, which is obtained as 0.9861, and
• the area to the left of -1.4 is obtained from the table in the intersection cell with row value -1.0 and column value 0.4, which is obtained as 0.0808.

Mathematically, it is expressed as shown below.

$$\begin{gathered} \text{Area}\text{to}\text{the}\text{left}\text{of}2.2=P\left(Z<2.2\right) \\ =0.9861 \\ \text{Area}\text{to}\text{the}\text{left}\text{of}-1.4=P\left(Z<-1.4\right) \\ =0.0808 \end{gathered}$$

Substitute the values in equation (1).

$$\begin{gathered} \text{Area}\text{between}-1.4\text{and}2.2=0.9861-0.0808 \\ =0.9053 \end{gathered}$$

Therefore, the area of the shaded region is 0.9053.