Standard normal distribution, 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.

Greater than 0

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

The distribution of the bone density test score follows the standard normal distribution, and the variable for the bone density test score is denoted by Z.

Thus,

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

Steps to draw a normal curve:

1. Make a horizontal axis and a vertical axis.
2. Mark the points -3.5, -3.0, -2.5 up to 3.0 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 greater than 0.

The shaded area of the graph indicates the probability that the z-score is greater than 0.

Due to the one-to-one correspondence of the area and probability in the standard normal curve, the cumulative probability of 0 is the same as the area to the left of 0.

Also, the area to the right of 0 is equal to 1 minus the area to the left of 0.

The probability that the bone density test score is greater than 0 is computed as

$$\begin{gathered} P\left(Z>0\right)=\text{Area}\text{to}\text{right}\text{of}0 \\ =1-\text{Area}\text{to}\text{the}\text{left}\text{of}0 \\ =1-P\left(Z<0\right) \end{gathered}$$

Referring to the standard normal table for the negative z-score, the cumulative probability of 0 is obtained from the cell intersection for row -0.00 and the column value of 0.00, which is 0.5000.

$$\begin{gathered} P\left(Z>0\right)=1-0.5000 \\ =0.5000 \end{gathered}$$

Thus, the probability of the bone density test score greater than 0 is 0.5000.