Dr Sham Mahmood

Hello, thank you for your question Qainat. In order to understand the use of standard deviations in this question, it’s important to cover the use of standard deviations in osteoporosis.

The Z-score is very similar to the T-score, the difference with the Z-score is that the T-score compares your bone density to the average bone density of young healthy adults of your same gender.

Both T and Z-scores are expressed in standard deviations above and below the mean. The chart below will help you determine what your T score means.

So we don’t have a value for the t and z-score, but what we do have is the information that this patients result is less than -2 SDs away from the mean.

Using this information we know that 95.4% of the population lie between +2 and -2 SD away from the mean. Leaving 4.6% of the population having a bone mass density of either less than -2 or above +2.

In our question, we are focused on the patient with a lower BMD than the patient so we take the 4.6 and divide by 2, which gives us the answer of 2.3%.

Hope this helps.



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