91Ó°ÊÓ

Value,$x=948$

Mean,$\mathrm{μ}=944$

$z−$ score,$z=0.50$

The formula used:$z=\frac{(x-\mathrm{μ})}{\mathrm{σ}}$

$z−$ 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:

$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}$

Substitute values,

$0.50=\frac{948−944}{\mathrm{σ}}$

That becomes

$0.50=\frac{4}{\mathrm{σ}}$

That further becomes

$\mathrm{σ}=\frac{4}{0.50}=8$

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.