Reading off an X value from its z-score

Question

A quantity is built from a standard normal Z \sim N(0,1) by X = 5Z + 20, so X \sim N(20, 25). What value of X corresponds to a z-score of -1.5?

Accepted Answer

Standardizing goes Z = (X - \mu)/\sigma; running it in reverse recovers the raw value: X = \mu + z\,\sigma. From X = 5Z + 20 we read off \mu = 20 and \sigma = 5 (since the multiplier on Z is the standard deviation, and its square 25 is the variance). For z = -1.5: X = 20 + (-1.5)(5) = 20 - 7.5 = 12.5. So a value 1.5 standard deviations below the mean is 12.5. Takeaway. Reverse standardization is X = \mu + z\sigma; with \mu = 20, \sigma = 5, a z-score of -1.5 maps to X = 12.5.

Reading off an X value from its z-score

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