Standardizing vs Z-Score

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.

A z-score rescales one value into a count of standard deviations. For a population it is $z = \frac{x - \mu}{\sigma}$, where $\mu$ (mu) is the mean and $\sigma$ (sigma) is the standard deviation; for a sample it is $z = \frac{x - \bar{x}}{s}$, with $\bar{x}$ (x-bar) and $s$ in those roles. The units cancel in the division, so a z-score has no units of its own and values from different scales become comparable. The sign carries the direction and the size carries the distance.

A student scores 88 on a test with mean 80 and standard deviation 4, then 92 on a test with mean 86 and standard deviation 8. The raw scores say the 92 was the better performance. The z-scores say otherwise: $z = (88 - 80)/4 = 2.00$ against $z = (92 - 86)/8 = 0.75$. Relative to the class each score came from, the 88 is much the stronger result, and that comparison is the whole reason to standardize.

The wrong sentence to watch for is "her z-score of 1.5 means she beat 93 percent of the class." That 93.3 percent is the area to the left of 1.5 under the standard normal curve, so it is a fact about the normal model rather than about z-scores. A z-score is defined for any distribution that has a mean and a standard deviation, including badly skewed ones, and only when the distribution is roughly normal does the z-table turn it into a percentile.

Standardizing cannot repair shape. Subtracting and dividing by constants shifts and stretches the values without reordering them, so a right-skewed set of z-scores is still right-skewed and its outliers are still outliers.

Small samples put a hard ceiling on the answer. Using the sample standard deviation, no value among $n$ observations can have $|z|$ larger than $\frac{n-1}{\sqrt{n}}$. For $n = 5$ that ceiling is 1.79, so a z-score of 2 computed from five data points is an arithmetic error and not a rare event. The formula also breaks down when $s = 0$, which happens only when every value is the same.