91影视

The bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1.

As the distribution of bone density follows a standard normal distribution, the random variable for bone density is expressed as Z.

Thus,

$$\begin{gathered} Z~N\left(渭,{蟽}^2\right) \\ ~N\left(0,1^2\right) \end{gathered}$$

Sketch the standard normal curve as shown below, where the shaded regions represent the lowest 3% and highest 3% areas under the z-score.

In the graph, represents the cutoff for the lowest 3% z-score ofbone density scores.

$$\begin{gathered} {\text{Area to the left ofz}}_{\text{1}}=P\left(Z<z_1\right) \\ =0.03 \end{gathered}$$

In the standard normal table, the value closest to 0.03 is 0.0301, and the corresponding row value -1.8 and the column value 0.08 correspond to the z-score of -1.88.

Thus, the value of the cutoff score corresponding to the lowest 3% scores is -1.88.

In the graph, $z_2$represents the cutoff for the top 3% z-scores ofbone density scores.

$$\begin{gathered} {\text{Area to the right ofz}}_2=P\left(Z>z_2\right) \\ =0.03 \end{gathered}$$

The left-tailed area corresponding to the cutoff score is

$$\begin{gathered} P\left(Z>z_2\right)=0.03 \\ 1-P\left(Z<z_2\right)=0.03 \\ P\left(Z<z_2\right)=0.97 \end{gathered}$$

In the standard normal table, the value closest to 0.97 is 0.9699, and the corresponding row value 1.8 and the column value 0.08 correspond to the z-score of 1.88.

Thus, the value of the cutoff score corresponding to the top 3% scores is 1.88.

The shaded area of the graph indicates the probability that the z-score is lesser than -1.88 and greater than 1.88.

The two cutoff scores separating the bottom 3% and the top 3% scores of bone density are -1.88 and 1.88, respectively.