Question

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

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

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

Short answer

Part (a) J’s bone density in the hip is $1.45$ standard deviation less than the average bone density

Part (b) The standard deviation of bone density is $\sigma =5.5172$.

Step-by-step solution

Part (a) Step 1. Given

Judy's hip bone density is $948g/{cm}^2$

Judy's standardized bone density score is $z=-1.45$

The average hip bone density is $956g/{cm}^2$

Part (a) Step 2. Concept

The inflection points are the points where the curve shifts from being concave up to being concave down. These inflection points are always one standard deviation distant from the mean on a normal density curve.

Part (a) Step 3. Calculation

M's average bone density is $948g/{cm}^2$

G's average bone density is $944g/{cm}^2$

Judy's standardized score is negative, indicating that her bone density in the hip is $1.45$ standard deviations lower than the average bone density in the hip of women her age.

Part (b) Step 1. Calculation

Judy's results show that her hip has a bone density of $948g/{cm}^2$and a normalized score of $z=-1.45$. Judy's bone density in the hip is $=948g/{cm}^2$in the reference population of $25-$year-old women. The following formula is used to compute the standard deviation: $$\begin{gathered} z\left(Judy\right)=x-\mu /\sigma \\ \Rightarrow -1.45=948-956/\sigma \\ \Rightarrow \sigma =5.5172 \end{gathered}$$

Therefore, the standard deviation of bone density is σ = $5.5172.$