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.

Find P₁₀, the 10th percentile. This is the bone density score separating the bottom 10% from the top 90%.

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,

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

Let $P_{10}$be the z-score on the distribution of z that represents the 10th percentile.

It implies that the area to the left of $P_{10}$is equal to 0.10.

Mathematically,

$$\begin{gathered} {\text{area to the left ofP}}_{10}=P\left(Z<P_{10}\right) \\ =0.10 \end{gathered}$$

Sketch the standard normal curve as shown below, where the shaded region represents 10% area under the z-score $P_{10}$. Here, $P_{10}$represents the 10^th percentile.

In the standard normal table, the value closest to 0.10 is 0.1003, and the corresponding row value -1.2 and the column value 0.08 represent the z-score of -1.28.

Therefore,$P\left(Z<-1.28\right)=0.10$ , which implies that theshaded area of the graph that has an area of 0.10 corresponds to a z-score lesser than -1.28.

Thus,$P_{10}$, the 10th percentile is -1.28.