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.

Between -1.00 and 5.00

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 scores 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 -4,-3, -2, -1 up to 6 on the horizontal axis and points 0, 0.05,

0.10 up to 0.50 on the vertical axis.

1. Provide titles to the horizontal and vertical axes as z and P(z), respectively.
2. Shade the region between -1.00 and 5.00.

The shaded area of the graph indicates the probability of the z-score between -1.00 and 5.00.

As there is a one-to-one correspondence between the area and probabilities, the probability that the bone density between -1.00 and 5.00 is computed as

$P\left(-1.00<Z<5.00\right)=P\left(Z<5.00\right)-P\left(Z<-1.00\right) ...\left(1\right)$

Referring to the standard normal table for the negative z-score, the cumulative probability of 5 is obtained from the cell intersection for the row 3.50 and the column value 0.00, which is 0.9999.

Referring to the standard normal table for the negative z-score, the cumulative probability of -1 is obtained from the cell intersection for row -1 and the column value of 0.00, which is 0.1587.

Thus,

.$$\begin{gathered} P\left(Z<5.00\right)=0.9999 \\ P\left(Z<-1.00\right)=0.1587 \end{gathered}$$

The probability that the bone density between -1.00 and 5.00 is obtained by substituting the cumulative probability in equation (1) is

$$\begin{gathered} P\left(-1.00<Z<5.00\right)=P\left(Z<5.00\right)-P\left(Z<-1.00\right) \\ =0.9999-0.1587 \\ =0.8412 \end{gathered}$$

Thus, the probability of the bone density test score between -1.00 and 5.00 is 0.8412.