Question

Comparing bone density Refer to the previous exercise. One of Judy’s friends, Mary, has 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—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 13(b)? Are you surprised?

Short answer

Part (a) M's bones are healthier than J's bones.

Part (b) The standard deviation of the bone density in M’s reference population is 8

Step-by-step solution

Part (a) Step 1. Given

Mary’s bone density in the hip = $948g/{cm}^2$

Mary’s standardized score of bone density, $z=0.5$

The mean bone density in the hip =$944g/{cm}^2$

Part (a) Step 2. Concept

Formula used:

$z=\frac{x-mean}{s\tan darddeviation}$

Part (a) Step 3. Explanation

Since J's standardized score is negative, she has $1.45$ standard deviation less bone density in her hip than the average bone density in the hip of $25$-year-old women like J, while M's standardized score is positive, she has $0.50$ standard deviation more bone density in the hip than the average bone density in the hip of $35$-year-old women like M. M's bones are thus in better condition than Judy's. Therefore, M's bones are healthier than J's bones.

Part (b) Step 1. Calculation

M's results show a $948g/{cm}^2$ bone density in the hip and a normalised score of $z=0.50$. The hip bone density in the reference population of 35-year-old women like M is$944g/{cm}^2$. The following formula is used to compute the standard deviation:

$$\begin{gathered} z\left(M\right)=x-\mu /\sigma \\ 0.50=948-944/\sigma \\ \sigma =4/0.50 \\ \sigma =8 \end{gathered}$$

J's results show a $948g/{cm}^2$ bone density in the hip and a normalised score of $z=-1.45$. The hip bone density in the reference population of $25$-year-old women like J is $956g/{cm}^2$. The following formula is used to compute the standard deviation:

$$\begin{gathered} z\left(J\right)=x-\mu /\sigma \\ -1.45=948-956/\sigma \\ \sigma =-8/-1.45 \\ \sigma \approx 5.52 \end{gathered}$$

The elder age group has an $8g/{cm}^2$ standard deviation. This is a higher standard deviation than the $5.52g/{cm}^2$ of J's reference population. This makes sense because, unsurprisingly, the older women get, the more diversity there will be when comparing their bone density. In M's reference population, the standard variation of bone density is $8$.