Assume that a random selected subject is given a bone density test. Those scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find the bone density test score that can be used as cutoff values separating the lowest 18% indicating levels that are too low or too high.
To find the cutoff values separating the lowest 18% of bone density test scores, we need to determine the z-scores corresponding to the given percentile. In a standard normal distribution (mean = 0, standard deviation = 1), we can use z-scores to find percentiles.
The z-score corresponding to the lower 18th percentile can be found using a standard normal distribution table or a statistical calculator. Looking up the z-score for the lower 18th percentile, we find it to be approximately -0.924.
Similarly, to find the z-score corresponding to the upper 18th percentile (the highest values), we subtract the z-score of the lower 18th percentile from zero, resulting in 0.924.
Now, we can calculate the corresponding bone density test scores (cutoff values) using the z-scores and the formula:
x = (z * σ) + μ
where x represents the bone density test score, σ is the standard deviation (1 in this case), and μ is the mean (0).
For the lower cutoff value:
x1 = (-0.924 * 1) + 0 = -0.924
For the upper cutoff value:
x2 = (0.924 * 1) + 0 = 0.924
Graphically, this can be represented as follows:
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--------x2--------------
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x1 -(lower cutoff value) 0
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The bone density test score below -0.924 indicates levels that are too low, whereas the score above 0.924 indicates levels that are too high, based on the given percentile interval.