91影视

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,

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

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,

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

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,

$0.0228脳100=2.28\%$

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 鈥�2 is obtained from the cell intersection for rows 鈥�2.0 and the column value 0.00, which is 0.0228.

Expressing the result as a percentage,

$0.0228脳100=2.28\%$

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,

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

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 鈥�2 is obtained from the cell intersection for rows 鈥�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,

$0.9544脳100=95.44\%$

Thus, 95.44% of the scores are not significant.