Final answer:
The probability of a bone density test score greater than 0.81 would depend on the test's mean and standard deviation; we convert the score to a z-score and reference a standard normal distribution table to find the probability.
Step-by-step explanation:
To find the probability of a bone density test score greater than 0.81, we need more information about the mean (μ) and the standard deviation (σ) of the test scores. However, let's assume we are dealing with a different scenario mentioned in the provided information where test scores on an exam are normally distributed with a mean (μ) of 81 and a standard deviation (σ) of 15. We could use that information as an example to show how to find the probability for a test score greater than a specific value, such as 0.81 standard deviations above the mean.
Firstly, we convert the test score to a z-score, which is done by subtracting the mean from the score and then dividing by the standard deviation: z = (score - μ) / σ. In our given example, the z-score representing 0.81 standard deviations above the mean would just be 0.81 since the standard deviation is 15 points.
To find the probability that corresponds to this z-score, we would look up the value in a standard normal distribution table, use a calculator, or software that provides the cumulative probability for a standard normal distribution. This gives us the area to the left of the z-score. To find the area to the right, we subtract this value from one.
If we were looking up the z-score of 0.81 on a standard normal distribution table, we'd find the probability to the left of z and then calculate 1 - P(z ≤ 0.81) to find the desired probability of a test score greater than 0.81 standard deviations above the mean. If we used the z-score 2.326 as an example (which corresponds to the top 1% of the distribution), the probability of a test score higher than that would be the area to the right of z, which is 0.01.