WEBVTT

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Here is the problem: A score of 62 comes from a distribution with mean 75 and standard deviation 8. Find its z-score.

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Z-score from x, mean and SD.

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Subtract the mean from the value.

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Divide by the standard deviation.

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Look up the area to the left (percentile).

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So the answer is z equals negative 1.625 (percentile approximately 5.208 percent).

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62 lies 1.625 standard deviations below the mean; about 5.208 percent of a normal population lies below it.

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Reverse: μ plus zσ equals 75 plus (negative 1.625)(8) equals 62.