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 thatare obtained by using the standard normal tablefor zscores.

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 zscore.

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 score is $x=91$.

The corresponding z score is computed as

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{91-100}{15} \\ =-0.6 \end{gathered}$$

Therefore, the z score is -0.6.

The shaded area is represented as

$$\begin{gathered} P\left(X>91\right)=P\left(Z>-0.6\right) \\ =1-P\left(Z<-0.6\right) \end{gathered}$$

Referring to the standard normal table, the cumulative probability ofis obtained from the cell intersection for rows -0.6 and 0.0, which is 0.2743.

$$\begin{gathered} P\left(X>91\right)=1-0.2743 \\ =0.7257 \end{gathered}$$

Therefore, the area of the shaded region is 0.7257.