In Exercises 9–12, find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

A shaded region is shown in the graph for the standard normal distribution of bone density scores.

Any curve with a total area equal to 1 has a one-to-one relationship of the area under the curve with the probability.

The left-tailed area is equal to cumulative probabilities, which can be obtained using the standard normal table for z-scores.

In the case of right-tailed areas, the difference of these cumulative probabilities from 1 gives the required area toward the right of the z-score.

It is required to compute the area between the two z-scores-0.84 and 1.28.

Mathematically, you can say the following:

$$\begin{gathered} \text{Area}\text{between}-0.84\text{and}1.28=\text{Area}\text{to}\text{the}\text{left}\text{of}1.28-\text{Area}\text{to}\text{the}\text{left}\text{of}-0.84 \\ =P\left(Z<1.28\right)-P\left(Z<-0.84\right)...\left(1\right) \end{gathered}$$

By using the standard normal table,

• the area to the left of 1.28 is obtained from the table in the intersection cell with row value 1.2 and the column value 0.08, which is obtained as 0.8997.
• the area to the left of -0.84 is obtained from the table in the intersection cell with row value -0.8 and the column value 0.04, which is obtained as 0.2005.

Mathematically, it is expressed as follows:

$$\begin{gathered} \text{Area}\text{to}\text{the}\text{left}\text{of}1.28=P\left(Z<1.28\right) \\ =0.8997 \\ \text{Area}\text{to}\text{the}\text{left}\text{of}-0.84=P\left(Z<-0.84\right) \\ =0.2005 \end{gathered}$$

Substitute the values into equation (1).

$$\begin{gathered} \text{Area}\text{between}-0.84\text{and}1.28=0.8997-0.2005 \\ =0.6992 \end{gathered}$$

Thus, the shaded area between -0.84 and 1.28 is equal to 0.6992.