91Ó°ÊÓ

It is given that the population of the z-scores corresponding to the bone density test results is normally distributed with a mean value equal to 0 and a standard deviation equal to 1.

Let X be the bone density test score.

Then,

$$\begin{gathered} X~N\left(\mathrm{μ},{\mathrm{σ}}^2\right) \\ ~N\left(0,1^2\right) \end{gathered}$$

The formula of z-score is given as

$z=\frac{x-\mathrm{μ}}{\mathrm{σ}}$

It is given that $\mathrm{μ}=0\text{and}\mathrm{σ}=1$.So, the z-score is given as

$z=x$

Here, the required probabilities are obtained by using the standard normal table.

a.

The probability that the bone density test score is less than 1.54 is computed below.

$P\left(z<1.54\right)=0.9382$

Therefore, the probability that the bone density test score is less than 1.54 is equal to 0.9382.

b.

The probability that the bone density test score is greater than -1.54 is computed below.

$$\begin{gathered} P\left(z>-1.54\right)=1-P\left(z<-1.54\right) \\ =1-0.0618 \\ =0.9382 \end{gathered}$$

Therefore, the probability that the bone density test score is greater than-1.54 is equal to 0.9382.

c.

The probability that the bone density test score lies between -1.33 and 2.33 is computed below.

$$\begin{gathered} P\left(-1.33<z<2.33\right)=P\left(z<2.33\right)-P\left(z<-1.33\right) \\ =0.9901-0.0918 \\ =0.8983 \end{gathered}$$

Therefore, the probability that the bone density test score lies between -1.33 and 2.33 is equal to 0.8983.

d.

It is known that 25% of all the values are less than or equal to the first quartile.

In terms of standard normal probability, the first quartile can be expressed as

$$\begin{gathered} Q_1=P\left(Zâ\copyright ½z\right) \\ =0.25 \end{gathered}$$

The value of the z-score corresponding to the left-tailed probability of 0.25 is equal to -0.67.

Thus, the value of the first quartile, which separates the bottom 25% and the top 75% of the bone density test score, is equal to -0.67.

e.

It is known that the sample mean follows the normal distribution with a mean equal to ${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$ and a standard deviation equal to ${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$.

Here, $\mathrm{μ}$ refers to the population mean and has a value equal to 0, and $\mathrm{σ}$ refers to the population standard deviation and has a value equal to 1.

The sample size (n) is equal to 9.

The probability of the mean bone density score to be greater than 0.5 is computed below.

$$\begin{gathered} P\left(\stackrel{¯}{x}>0.5\right)=P\left(\frac{\stackrel{¯}{x}-\mathrm{μ}}{\frac{\mathrm{σ}}{\sqrt{n}}}>\frac{0.5-\mathrm{μ}}{\frac{\mathrm{σ}}{\sqrt{n}}}\right) \\ =P\left(z>\frac{0.5-0}{\frac{1}{\sqrt{9}}}\right) \\ =P\left(z>1.5\right) \\ =1-P\left(z<1.5\right) \end{gathered}$$

Thus, the probability of the mean bone density score being greater than 0.5 is equal to 0.0668.