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%

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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.
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