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 1.28

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,

Steps to sketch a normal curve:

• Make a horizontal and a vertical axis.
• 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.
• Provide titles to the horizontal and vertical axes as z and P(z), respectively.
• Shade the region lesser than 1.28.

The shaded area of the graph indicates the probability that the z-score is lesser than 1.28. Due to the one-to-one correspondence of the area and probability in the standard normal curve, the cumulative probability of 1.28 is the same as the area to the left of 1.28.

Referring to the standard normal table for the positive z-score, the cumulative probability of 1.28 is obtained from the cell intersection for row 1.2 and the column value 0.08, which is 0.8997.

The probability that the bone density is lesser than 1.28 is computed as

$$\begin{gathered} \text{Area to the left of 1.28}=P\left(z<1.28\right) \\ =0.8997 \end{gathered}$$

Thus, the probability of the bone density test score being less than 1.28 is 0.8997.