To determine the percentage of the older population aged 70 to 79 that has osteoporosis, we can use the concept of standard deviations and Z-scores.
1. Given that the mean bone mineral density (BMD) in this age group is about -2 on the standard scale for young adults and the standard deviation is the same as for young adults, we have a mean (μ) of -2 and a standard deviation (σ) of 1 (assuming a standard normal distribution where σ = 1).
2. Osteoporosis is typically defined as having a BMD that is 2.5 standard deviations below the mean or lower. In this case, since the mean BMD for the older population is -2, osteoporosis would be indicated by a BMD of -2 - (2.5 * 1) = -2 - 2.5 = -4.5.
3. To find the percentage of the older population with osteoporosis, we can calculate the Z-score for a BMD of -4.5 using the formula Z = (X - μ) / σ, where X is the value we are interested in.
4. Plugging in the values, we get Z = (-4.5 - (-2)) / 1 = -2.5. This means that a BMD of -4.5 corresponds to a Z-score of -2.5.
5. Using a standard normal distribution table or a calculator, we can find the percentage of values below a Z-score of -2.5, which corresponds to the percentage of the older population with osteoporosis.
6. By looking up the Z-score of -2.5 in a standard normal distribution table, we find that approximately 0.0062 (or 0.62%) of the population falls below this value.
Therefore, about 0.62% of the older population aged 70 to 79 has osteoporosis.
