Standard normal distribution. In Exercise 17-36, assume that a randomly selected subject is given a bone density test. Those 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 probability of the given bone density test score. If using technology instead of Table A-2, round answers to four decimal places.

Less than -1.23

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,

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

Steps to sketch a normal curve:

1. Make a horizontal and a vertical axis.
2. Mark the points -3.0, -2.5, -2.0 up to 3 on the horizontal axis and points 0, 0.05, 0.10 up to 0.50 on the vertical axis.
3. Provide titles to the horizontal and vertical axes as z and P(z), respectively.
4. Shade the region lesser than -1.23.

The shaded area of the graph indicates the probability that the z-score is lesser than -1.23. Due to the one-to-one correspondence of the area and probability in the standard normal curve, the cumulative probability of -1.23 is the same as the area to the left of 1.23.

Referring to the standard normal table, the cumulative probability of -1.23 is obtained from the cell intersection for rows -1.2 and the column value 0.03, which is 0.1093.

The probability that the bone density is lesser than -1.23 is computed as follows.

$$\begin{gathered} \text{Area to the left of}-1.23=P\left(z<-1.23\right) \\ =0.1093 \end{gathered}$$

Thus, the probability of the bone density test score being less than -1.23 is 0.1093.