Significance For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that are

a. significantly high (or at least 2 standard deviations above the mean).

b. significantly low (or at least 2 standard deviations below the mean).

c. not significant (or less than 2 standard deviations away from the mean).

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

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

Thus,

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

a.

The percentage of scores that are significantly high are the least 2 standard deviations above the mean.

Thus, the probability to fall beyond 2 standard deviations to the right of the mean is,

$P\left(Z>2\right)=1-P\left(Z<2\right) ...\left(1\right)$

Referring to the standard normal table, the cumulative probability of 2 is obtained from the cell intersection for rows 2.0 and the column value 0.00,which is 0.9772.

Substitute the cumulative probability in equation (1) as,

$$\begin{gathered} P\left(Z>2\right)=1-P\left(Z<2\right) \\ =1-0.9772 \\ =0.0228 \end{gathered}$$

Expressing the result as a percentage,

$0.0228\times 100=2.28\%$

Thus, 2.28% of the scores are significantly high.

b.

The percentage of the scores that are significantly low arethe least 2 standard deviations below the mean.

Thus, the probability to fall beyond 2 standard deviations to the left of the mean is,$P\left(Z<-2\right)$.

Referring to the standard normal table, the cumulative probability of –2 is obtained from the cell intersection for rows –2.0 and the column value 0.00, which is 0.0228.

Expressing the result as a percentage,

$0.0228\times 100=2.28\%$

Thus, 2.28% scores are significantly low.

c.

The percentage of the scores that are not significant fall in2 standard deviations range of the mean, andexpressed as,

$P\left(-2<Z<2\right)=P\left(Z<2\right)-P\left(Z<-2\right) ...\left(2\right)$

Referring to the standard normal table,

• the cumulative probability of 2 is obtained from the cell intersection for rows 2.0 and the column value 0.00, which is 0.9772.
• the cumulative probability of –2 is obtained from the cell intersection for rows –2.0 and the column value 0.00, which is 0.0228.

Substitute the cumulative probabilities in equation (2) as,

$$\begin{gathered} P\left(-2<Z<2\right)=P\left(Z<2\right)-P\left(Z<-2\right) \\ =0.9772-0.0228 \\ =0.9544 \end{gathered}$$

Expressing the result as a percentage,

$0.9544\times 100=95.44\%$

Thus, 95.44% of the scores are not significant.