Question

Finding Bone Density Scores. In Exercises 37–40 assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the bone density test score corresponding to the given information. Round results to two decimal places.

If bone density scores in the bottom 2% and the top 2% are used as cutoff points for levels that are too low or too high, find the two readings that are cutoff values.

Short answer

The graph for two cutoff scores of bone density scores in the bottom 2% and the top 2% region is shown below.

The cutoff value for the bone density scores in the bottom 2% is -2.05, and the cutoff value for the bone density scores in the top 2% is 2.05.

Step-by-step solution

Given information

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

Describe the distribution

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

Thus,

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

Draw a graph for the bone density scores 

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.

Find the cutoff for the bottom 2% scores

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.

Find the cutoff for the top 2% 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.

Summarize the result

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.