Standardizing

Standardizing rewrites a value as the number of standard deviations it sits from the mean. With population parameters it is $z = \frac{x - \mu}{\sigma}$, where $\mu$ (mu) is the mean and $\sigma$ (sigma) is the standard deviation. When those are unknown the sample versions stand in, $z = \frac{x - \bar{x}}{s}$. The result carries no units, and that is the whole point of doing it.

A reading score of 640 on a test with mean 500 and standard deviation 100 standardizes to $z = \frac{640 - 500}{100} = 1.40$. A math score of 27 on a test with mean 21 and standard deviation 5 gives $z = \frac{27 - 21}{5} = 1.20$. The raw numbers are not comparable and the z-scores are, so the reading result is the stronger of the two against its own test.

"Standardize the data and you have made it normal." Standardizing subtracts a constant and divides by a constant. That slides a distribution along and rescales it; it cannot bend it. Standardize 1, 2, 3, 4, 20, whose mean is 6 and whose $s$ is 7.91, and you get $-0.63$, $-0.51$, $-0.38$, $-0.25$, and $1.77$: still four values bunched together with one stranded far to the right. The mean sat 3 units above the median before, and 0.38 standard deviations above it after, which is the same gap in new clothing.

What standardizing does guarantee is arithmetic rather than shape. As long as the standard deviation is not 0, a standardized data set has mean exactly 0 and standard deviation exactly 1, so z-scores that fail to average to 0 mean you subtracted the wrong mean. Data with every value identical is the one case where the operation is undefined, since $s = 0$ puts a zero in the denominator.

Topic 1.9 of the Fall 2026 course asks you to calculate z-scores with population parameters and notes that the sample mean and standard deviation may be used when the population values are unknown. The same move produces the standardized test statistic later on, where the divisor becomes a standard error.