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 -3.00 and 3.00.

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=-3.00 and z=3.00.

The shaded area represents the probability.

Using table A-2,

The probability that the bone density score is between -3.00 and 3.00 is computed as follows.

$$\begin{gathered} P\left(-2.00<z<2.00\right)=P\left(z<3.00\right)-P\left(z<-3.00\right) \\ =0.9987-\left(1-P\left(z<3.00\right)\right) \\ =0.9987-\left(1-0.9987\right) \\ =0.9974 \end{gathered}$$

Thus, the probability that the bone density score is between -3.00 and 3.00 is 0.9974.