Measuring bone density Individuals with low bone density (osteoporosis) have a high risk of broken bones (fractures). Physicians who are concerned about low bone density in patients can refer them for specialized testing. Currently, the most common method for testing bone density is dual-energy $X-$ray absorptiometry (DEXA). The bone density results for a patient who undergoes a DEXA test usually are reported in grams per square centimeter ($g/{cm}^2$) and in standardized units. Judy, who is $25$years old, has her bone density measured using DEXA. Her results indicate bone density in the hip of 948 g/cm2 and a standardized score of $z=-1.45$The mean bone density in the hip is 956 $g/{cm}^2$in the reference population of $25-$year-old women like Judy.

a. Judy has not taken a statistics class in a few years. Explain to her in simple language what the standardized score reveals about her bone density.

b. Use the information provided to calculate the standard deviation of bone density in the reference population.

Step by step solution

01

Part (a) Step 1: Given information

Value, $x=948$

Mean,$\mu =956$

z − score, $z=-1.45$

02

Part (a) Step 2: Concept

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

03

Part (a) Step 3: Explanation

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.

Thus,

Judy's bone density of $948g/{cm}^2$ is $1.45$ standard deviations below the mean bone density of $25-$year-old women, according to the standardized $z-$ score.

04

Part (b) Step 1: Calculation

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

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

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

Thus,

The standard deviation is approx. $5.52g/{cm}^2$

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