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

As the distribution of the bone density follows the standard normal distribution, the random variable for bone density is expressed as Z.

Thus,

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

Steps to make a normal curve:

• Make a horizontal and a vertical axis.
• Mark the points -3.0, -2.5, -3.0 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 2.56.

The shaded area of the graph indicates the probability that the z-score is lesser than 2.56. Due to the one-to-one correspondence of the area and probability in the standard normal curve, the cumulative probability of 2.56 is the same as the area to the left of 2.56.

Referring to the standard normal table for the positive z-score, the cumulative probability of 2.56 is obtained from the cell intersection for row 2.5 and the column value 0.06, which is 0.9948.

The probability that the bone density is lesser than 2.56 is computed as

$$\begin{gathered} \text{Area to the left of 2.56}=P\left(z<2.56\right) \\ =0.9948 \end{gathered}$$

Thus, the probability of the bone density test score being less than 2.56 is 0.9948.