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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 0.9918.

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 $\mathrm{μ}=100$, and the standard deviation is $\mathrm{σ}=15$.

The area to the left of the value x with the corresponding z score z is 0.9918.

Mathematically,

$$\begin{gathered} P\left(X<x\right)=P\left(Z<z\right) \\ =0.9918 \end{gathered}$$

Using the standard normal table, the area of 0.9918 is observed corresponding to the row value 2.4 and the column value 0.00. This implies that the z score is 2.4.

Therefore,

$P\left(Z<2.4\right)=0.9918$

Thus, the z score corresponding to x is 2.4.

The IQ score is computed as

$$\begin{gathered} \frac{x-\mathrm{μ}}{\mathrm{σ}}=2.4 \\ \frac{x-100}{15}=2.4 \\ x=\left(2.4×15\right)+100 \\ =136 \end{gathered}$$

Therefore, the value of x is 136.

Thus, the IQ score that has the left-tailed area as 0.9918 is 136.