In Exercises 9–12, find the indicated IQ score and round to the nearest whole number. 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 area of the shaded region is 0.9099.

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 area to the left of x is 0.9099, with the corresponding z score of z.

Mathematically,

.$P\left(Z<z\right)=0.9099$

Using the standard normal table, the area of 0.9099 is observed corresponding to the row value 1 and the column value 0.34. This implies that the z score is 1.34.

Therefore,

.$P\left(Z<1.34\right)=0.9099$

Thus, the z score is 1.34.

The IQ score is computed as

$$\begin{gathered} \frac{x-\mu }{\sigma }=1.34 \\ \frac{x-100}{15}=1.34 \\ x=\left(1.34\times 15\right)+100 \\ \approx 120 \end{gathered}$$

Thus, the IQ score x, the area to the left of which is 0.9099, is 120.