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Here is the problem: A score of 130 comes from a distribution with mean 100 and standard deviation 15. 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 2 (percentile approximately 97.72 percent).

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130 lies 2 standard deviations above the mean; about 97.72 percent of a normal population lies below it.

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Reverse: μ plus zσ equals 100 plus (2)(15) equals 130.