39-score of a normal value
. Find the -score of .
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.
. Find the -score of .
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.