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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 =$944g/{cm}^2$

Formula used:

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

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.

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:

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:

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 $8$.