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 graph representing the IQ scores of adults is normally distributed.

The mean is 100.

The standard deviation is 15.

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 $x=118$.

The shaded area is the left-tailed area on the distribution curve to the value 118.

As the area under the normal curve is 1, there is a one-to-one correspondence between probabilities and area under the curve.

The corresponding z score is computed as

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{118-100}{15} \\ =1.2 \end{gathered}$$

Therefore, the z score is 1.2.

The shaded area is represented as

.$P\left(X<118\right)=P\left(Z<1.2\right)$

Referring to the standard normal table, the probability is obtained from the intersection cell of row 1.2 and column 0.0,which is 0.8849.

Therefore, the area of the shaded region is 0.8849.