91Ó°ÊÓ

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

Let Z be the random variable for bone mineral density test scores.

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

The probability of a randomly selected score being greater than –2.93 is expressed as,

$P\left(Z>-2.93\right)$

Since the probability has one-to-one correspondence with the area under the curve, the expression infers to the right tailed area of –2.93.

Thus, the area to the right of –2.93 is 1 minus the area to the left of –2.93.

Mathematically,

$$\begin{gathered} P\left(Z>-2.93\right)=1-\text{Area}\text{to}\text{the}\text{left}\text{of}-2.93 \\ =1-P\left(Z<-2.93\right)...\left(1\right) \end{gathered}$$

Using the standard normal table, the cumulative probability of corresponding to intersection of row –2.9 and column 0.03 is 0.0017.

Thus, $P\left(Z<-2.93\right)=0.0017$.

Substituting in equation (1),

$$\begin{gathered} P\left(Z>-2.93\right)=1-0.0017 \\ =0.9983 \end{gathered}$$

Therefore, the probability of any randomly selected score being greater than –2.93 is 0.9983.