91Ó°ÊÓ

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 the standard normal distribution, the random variable for bone density is expressed as Z.

Thus,

$$\begin{gathered} Z~N\left(\mathrm{μ},{\mathrm{σ}}^2\right) \\ ~N\left(0,1^2\right) \end{gathered}$$

Let $P_{99}$be the z-score on the distribution of z that represents the 99th percentile.

It implies that the area to the left of $P_{99}$is equal to 0.99.

Mathematically,

$$\begin{gathered} {\text{Area to the left ofP}}_{99}=P\left(Z<P_{99}\right) \\ =0.99 \end{gathered}$$

Sketch the standard normal curve, as shown below, where the shaded region represents 99% area under the z-score $P_{99}$. Here, $P_{99}$represents the 99^th percentile.

In the standard normal table, find the value closest to 0.99, which is 0.9901. The corresponding row value is 2.3, and the column value is 0.03. This corresponds to the z-score of2.33, which is the value of $P_{99}$.

The shaded area of the graph indicates the probability that the z-score is lesser than 2.33.

Thus, $P_{99}$the 99th percentile is 2.33.