Chapter 2.1, Problem 18E

(a)

To determine

Comparison of standardized scores for healthier bones.

Answer to Problem 18E

Mary’s bones are healthier for her age.

Explanation of Solution

Given information:

z − scores,

For Judy,

  $z=-1.45$

For Mary,

  $z=0.50$

The z − score describes the number of standard deviations that a value is from the mean.

Negative z − score indicates that the value is below the mean.

Whereas,

Positive z − score indicates that the value is above the mean.

That implies

Higher z − score is better (or healthier).

In the reference population, the higher z − score indicates the higher bone density.

Thus,

Mary’s bones are healthier for her age.

(b)

To determine

Standard deviation of the density in Mary’s reference population and compare this with Judy’s case.

Answer to Problem 18E

Standard deviation,

  $\sigma =8$

Standard deviation of the density is higher in Mary’s case.

Explanation of Solution

Given information:

Value, $x=948$

Mean, $\mu =944$

z − score, $z=0.50$

Calculate the z − score:

  $z=\frac{x-\mu }{\sigma }$

Substitute values,

  $0.50=\frac{948-944}{\sigma }$

That becomes

  $0.50=\frac{4}{\sigma }$

That further becomes

  $\sigma =\frac{4}{0.50}=8$

Since the units of the standard deviation and the data values are same.

Thus,

The standard deviation is 8 g/cm².

Surprisingly, standard deviation of the density in Mary’s case is higher than that of Judy’s case.