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.

A density curve with an aggregate area equal to 1 unit has a one-to-one association with theprobabilityunder the curve in a specific range.

The left-tailed areas are simply equal to cumulative probabilities, which can be obtained using the standard normal table 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.

It is required to compute the area between the two z-scores,-1.07 and 0.67.

Mathematically, you can see the following:

$$\begin{gathered} \text{Area}\text{between}-1.07\text{and}0.67=\text{Area}\text{to}\text{the}\text{left}\text{of}0.67-\text{Area}\text{to}\text{the}\text{left}\text{of}-1.07 \\ =P\left(Z<0.67\right)-P\left(Z<-1.07\right)...\left(1\right) \end{gathered}$$

By using the standard normal table,

• the area to the left of 0.67 is obtained from the table in the intersection cell with row value 0.6 and the column value 0.07, which is obtained as 0.7486.
• the area to the left of -1.07 is obtained from the table in the intersection cell with row value -1.0 and the column value 0.07, which is obtained as 0.1423.

Mathematically, it is expressed as follows:

$$\begin{gathered} \text{Area}\text{to}\text{the}\text{left}\text{of}0.67=P\left(Z<0.67\right) \\ =0.7486 \\ \text{Area}\text{to}\text{the}\text{left}\text{of}-1.07=P\left(Z<-1.07\right) \\ =0.1423 \end{gathered}$$

Substitute the values into equation (1).

$$\begin{gathered} \text{Area}\text{between}-1.07\text{and}0.67=0.7486-0.1423 \\ =0.6063 \end{gathered}$$

Thus, the shaded area between -1.07 and 0.67 is equal to 0.6063.