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.

Less than 4.55

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

The distribution of bone density follows the standard normal distribution, and the variable for bone density 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 up to 6 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 less than 4.55.

The shaded area of the graph indicates the probability that the z-score is lesser than 4.55.

Referring to the standard normal table for the negative z-score, the cumulative probability of 4.55 is obtained from the cell intersection for row 4.5 and the column value of 0.05, which is 0.9999.

As the probability and area have a one-to-one correspondence, the probability that the bone density test score is less than 4.55 is computed as

$$\begin{gathered} \text{Area}\text{to}\text{left}\text{of}4.55=P\left(Z<4.55\right) \\ =0.9999 \end{gathered}$$

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Thus, the probability of the bone density test score less than 4.55 is 0.9999.