Assume that a random selected subject is given a bone density test. Bone test score are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find P8 the 8th percentile. This is the bone density score separating the bottom 8% from the top 92%
To find the 8th percentile, we need to locate the z-score that corresponds to that percentile and then convert it back to the original bone density score using the mean and standard deviation.
First, let's find the z-score using a standard normal distribution table or calculator.
From the table or calculator, we find that the z-score corresponding to the 8th percentile is approximately -1.405.
Now, we can convert this z-score back to the original bone density score using the formula:
X = μ + (z * σ)
Where:
X = Bone density score
μ = Mean (0 in this case)
z = z-score (-1.405)
σ = Standard deviation (1 in this case)
Plugging in the values, we get:
X = 0 + (-1.405 * 1)
X = -1.405
Therefore, the bone density score separating the bottom 8% from the top 92% is approximately -1.405.