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,

\(\begin{aligned}{c}X \sim N\left( {\mu ,{\sigma ^2}} \right)\\ \sim N\left( {0,{1^2}} \right)\end{aligned}\).

The formula of z-score is given as

\(z = \frac{{x - \mu }}{\sigma }\).

It is given that\(\mu = 0\;{\rm{and}}\;\sigma = 1\).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.

\(\begin{aligned}{c}P\left( {z > - 1.54} \right) = 1 - P\left( {z < - 1.54} \right)\\ = 1 - 0.0618\\ = 0.9382\end{aligned}\).

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.

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

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

\(\begin{aligned}{c}{Q_1} = P\left( {Z \le z} \right)\\ = 0.25\end{aligned}\).

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 \({\mu _{\bar x}} = \mu \) and a standard deviation equal to \({\sigma _{\bar x}} = \frac{\sigma }{{\sqrt n }}\).

Here, \(\mu \) refers to the population mean and has a value equal to 0, and \(\sigma \) 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.

\(\begin{aligned}{c}P\left( {\bar x > 0.5} \right) = P\left( {\frac{{\bar x - \mu }}{{\frac{\sigma }{{\sqrt n }}}} > \frac{{0.5 - \mu }}{{\frac{\sigma }{{\sqrt n }}}}} \right)\\ = P\left( {z > \frac{{0.5 - 0}}{{\frac{1}{{\sqrt 9 }}}}} \right)\\ = P\left( {z > 1.5} \right)\\ = 1 - P\left( {z < 1.5} \right)\end{aligned}\)

\(\begin{aligned}{c} = 1 - 0.9332\\ = 0.0668\end{aligned}\)

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