91Ó°ÊÓ

The bone density test scores are normally distributed.

The mean score is $\mathrm{μ}=0$.

The standard deviation is$\mathrm{σ}=1$.

The z-scores are provided as -2.75 and -0.75.

Let x represent the bone density test score.

Asthe mean and standard deviation are0 and 1, respectively,x follows a standard normal distribution.

Steps to make a normal curve:

Step 1: Make a horizontal and a vertical axis.

Step 2: Mark the points -4, -2, 0, 2, and 4 on the horizontal axis and points 0.1, 0.2, 0.3, and 0.4 on the vertical axis.

Step 3: Provide titles to the horizontal and vertical axes as ‘z’ and ‘f(z)’, respectively.

Step 4: Shade the region between z=-2.75 and z=-0.75.

The shaded area represents the probability.

Using table A-2,

• the area to the left of 0.75is obtained from the table in the intersection cell with the row value 0.7 and the column value 0.05, which is obtained as 0.7734, and
• the area to the left of 2.75 is obtained from the table in the intersection cell with the row value 2.7 and the column value 0.05, which is obtained as 0.997.

The probability that the bone density score is between -2.75 and -0.75 is computed as follows.

$$\begin{gathered} P\left(-2.75<z<-0.75\right)=P\left(z<-0.75\right)-P\left(z<-2.75\right) \\ =\left(1-P\left(z<0.75\right)\right)-\left(1-P\left(z<2.75\right)\right) \\ =\left(1-0.7734\right)-\left(1-0.9970\right) \\ =0.2236 \end{gathered}$$

Thus, the probability that the bone density score is between -2.75 and -0.75 is 0.2236.