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.
Sham
