If X is a random variable from a normal distribution with mean μ and standard deviation σ, its Z-score may be calculated from X by subtracting μ and dividing by the standard deviation: If
is the total of a sample of size n from some population in which the mean is μ and the standard deviation is σ, the expected total is nμ and the standard error is
But since the normal distribution curve is symmetrical, probabilities for only positive values of Z are typically given.
The user might have to use a complementary operation on the absolute value of Z, as in the example below.
The values are calculated using the cumulative distribution function of a standard normal distribution with mean of zero and standard deviation of one, usually denoted with the capital Greek letter
Note that for z = 1, 2, 3, one obtains (after multiplying by 2 to account for the [−z,z] interval) the results f (z) = 0.6827, 0.9545, 0.9974, characteristic of the 68–95–99.7 rule.
[5] This table gives a probability that a statistic is greater than Z, for large integer Z values.
A professor's exam scores are approximately distributed normally with mean 80 and standard deviation 5.