Chapter 2.1, Problem 14E

(a)

To determine

To find: Whether J’s bones are healthier or M’s bone healthier

Answer to Problem 14E

M's bones are healthier than J's bones.

Explanation of Solution

Given:

Mean bone density of M = ${\text{948 g/cm}}^{\text{2}}$

Mean bone density of G = ${\text{944 g/cm}}^{\text{2}}$

Calculation:

Since J's standardized score is negative this means her bone density in hip is 1.45 standard deviation less than the average bone density in hip of population of 25-year-old women like J and M's standardized score is positive this means her bone density in hip is 0.50 standard deviation above the average bone density in hip of population of 35-year-old women like M. This means M's bones are healthier than Judy's bones.

Conclusion:

Therefore, M's bones are healthier than J's bones.

(b)

To determine

To find: the standard deviation of the bone density in M’s reference population.

Answer to Problem 14E

The standard deviation of the bone density in M’s reference population is 8

Explanation of Solution

Calculation:

M's results indicate a bone density in hip of $948g/c{m}^2$ and a standardized score of $z=0.50$ . In the reference population of 35-year-old women like M, the bone density in the hip is $944g/c{m}^2$ . The standard deviation is calculated below:

  $\begin{array}{l}z(\text{M})=\frac{x-\mu }{\sigma }\\ 0.50=\frac{948-944}{\sigma }\\ \sigma =\frac{4}{0.50}\\ \sigma =8\end{array}$

J's results indicate a bone density in hip of 948 g/cm2 and a standardized score of $z=-1.45$ . In the reference population of 25-year-old women like J, the bone density in the hip is $956g/c{m}^2$ . The standard deviation is calculated below:

  $\begin{array}{l}z(J)=\frac{x-\mu }{\sigma }\\ -1.45=\frac{948-956}{\sigma }\\ \sigma =\frac{-8}{-1.45}\\ \sigma \approx 5.52\end{array}$

The older age group has a standard deviation of $8g/c{m}^2$ . This is a larger standard deviation than J's reference population $\left(5.52g/c{m}^2\right).$ this does make sense because the older women get the more variability there will be when comparing their bone density, which is not surprising.

Conclusion:

The standard deviation of the bone density in M’s reference population is 8