Standard Normal Distribution. Find the indicated z score. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

The bone density score follows the standard normal distribution with a mean of 0 and a standard deviation of 1.

The z-score is the distance along the horizontal scale of the standard normal distribution (corresponding to the number of standard deviations above or below the mean). These are the values taken by the standard normal distribution, represented by Z.

As the standard normal curve has an area of 1 enclosed under the curve, the area has a one-to-one correspondence with the probability.

The left-tailed area is equal to the cumulative probabilities, which are obtained by using the standard normal table (Table A-2) for z-scores.

To find the required right-tailed area, subtract the cumulative probabilities from 1.

The area represented is mathematically represented as

$$\begin{gathered} \text{Area}\text{to}\text{the}\text{right}\text{of}\text{z}=P\left(Z>z\right) \\ 0.2061=1-P\left(Z<z\right) \\ P\left(Z<z\right)=1-0.2061 \\ P\left(Z<z\right)=0.7939 \end{gathered}$$

The z-score that has the area of 0.7939 to the left of it is obtained from the standard normal table as follows.

From the standard normal table, the area of 0.7939 is observed corresponding to the row value of 0.8 and the column value of 0.02, which implies that the z-score is 0.82.

Thus, the indicated z-score value with the right-tailed area of 0.2061 is 0.82.