silemus.

39zz-score of a normal value

X∼N(70,52)\displaystyle X \sim N(70, 5^2). Find the z\displaystyle z-score of x=80\displaystyle x = 80.

A z-score measures distance from the mean in units of standard deviation: z = (x − μ)/σ. Here z = (80 − 70)/5 = 2, so 80 is 2 standard deviations above the mean. Check: 70 + 2(5) = 80.

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