Comparing bone density Refer to Exercise $17$ Judy’s friend Mary also had the bone density in her hip measured using DEXA. Mary is $35$ years old. Her bone density is also reported as $948g/{cm}^2$ but her standardized score is $z=0.50$ The mean bone density in the hip for the reference population of 35-year-old women is $944grams/{cm}^2$

a. Whose bones are healthier for her age: Judy’s or Mary’s? Justify your answer.

b. Calculate the standard deviation of the bone density in Mary’s reference population. How does this compare with your answer to Exercise $17\left(b\right)$? Are you surprised?

Value,$x=948$

Mean,$\mu =944$

$z-$ score,$z=0.50$

The formula used:$z=\frac{(x-\mu )}{\sigma }$

$z-$ scores,

For Judy, $z=-1.45$

For Mary, $z=0.50$

The amount of standard deviations a value deviates from the mean is described by the $z-$score.

The value is below the mean when the $z-$score is negative.

Whereas,

Positive $z-$implies that the value is greater than the mean.

That implies

The higher the $z-$score, the better (or healthier).

A higher $z-$score in the reference population suggests higher bone density.

As a result, Mary’s bones are healthier for her age.

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

$\sigma =\frac{4}{0.50}=8$

Because the standard deviation and data values have the same units.

Thus,

The standard deviation is $8g/{cm}^2$

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