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 bottom 2% and top 2% areas under the z-scores on a standard normal curve.

In the graph, represents the z-score of thebone density scores in the bottom 2%.

Mathematically,

$$\begin{gathered} {\text{area to the left ofz}}_{\text{1}}=P\left(z<z_1\right) \\ =0.02 \end{gathered}$$

In the standard normal table, the value closest to 0.02 is 0.0202, and the corresponding row value -2.0 and the column value 0.05 correspond to the z-score of -2.05.

Thus,$P\left(Z<-2.05\right)=0.02$

Thus, the value of is equal to -2.05, which is the cutoff value for the bottom 2% bone density scores.

In the graph, $z_2$represents the cutoff z-score of thebone density score in the top 2%.

Mathematically,

$$\begin{gathered} P\left(Z>z_2\right)=0.02 \\ 1-P\left(Z<z_2\right)=0.02 \\ P\left(Z<z_2\right)=0.98 \end{gathered}$$

Thus, the cumulative area under the z-score is 0.98.

In the standard normal table, the value closest to 0.98 is 0.9798, and the corresponding row value 2.0 and the column value 0.05 correspond to the z-score of 2.05.

Thus, the value of $z_2$is equal to 2.05, which is the cutoff value for the top 2% bone density scores.

The shaded area of the graph indicates the probability that the z-score is lesser than -2.05 and greater than 2.05.

Thus, the two readings for the cutoff values are -2.05 and 2.05.