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/cm2, but her standardized score isz=0.50. The mean bone density in the hip for the reference population of 35-year-old women is 944grams/cm2.

(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

Expert verified

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

01

Part (a) Step 1. Given

Mary’s bone density in the hip = 948g/cm2

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

The mean bone density in the hip =944g/cm2

02

Part (a) Step 2. Concept

Formula used:

z=xmeanstandarddeviation

03

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.

04

Part (b) Step 1. Calculation

M's results show a 948g/cm2 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 is944g/cm2. The following formula is used to compute the standard deviation:

z(M)=xμ/σ0.50=948944/σσ=4/0.50σ=8

J's results show a 948g/cm2 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/cm2. The following formula is used to compute the standard deviation:

z(J)=xμ/σ1.45=948956/σσ=8/1.45σ5.52

The elder age group has an 8g/cm2 standard deviation. This is a higher standard deviation than the 5.52g/cm2 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.

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Most popular questions from this chapter

T2.1. Many professional schools require applicants to take a standardized test. Suppose that 1000 students take such a test. Several weeks after the test, Pete receives his score report: he got a 63, which placed him at the 73rd percentile. This means that

(a) Pete's score was below the median.

(b) Pete did worse than about 63%of the test takers.

(c) Pete did worse than about 73%of the test takers.

(d) Pete did better than about 63%of the test takers.

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