Chapter 2.1, Problem 17E

(a)

To determine

Interpretation of the standardized score in simple language.

Explanation of Solution

Given information:

z − score,

  $z=-1.45$

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.

Thus,

The standardized z − score reveals that Judy’s bone density of 948 g/cm² is 1.45 standard deviations below the mean bone density of 25 − year − old women.

(b)

To determine

.Standard deviation of bone density in the reference population using the information provided.

Answer to Problem 17E

Standard deviation,

  $\sigma =5.52$

Explanation of Solution

Given information:

Value, $x=948$

Mean, $\mu =956$

z − score, $z=-1.45$

Calculations:

Calculate the z − score:

  $z=\frac{x-\mu }{\sigma }$

Substitute values,

  $-1.45=\frac{948-956}{\sigma }$

That becomes

  $\sigma =\frac{-8}{-1.45}$

That further becomes

  $\sigma =\frac{8}{1.45}\approx 5.52$

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

Thus,

The standard deviation is approx. 5.52 g/cm².