Z-Score vs Standard Deviation

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.

Standard deviation reports spread as a typical distance between a value and the mean, carried in the units of the data. For a sample it is $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$, where $x_i$ are the observations, $\bar{x}$ (x-bar) is the sample mean, and $n$ is how many values there are. For a whole population the symbol becomes $\sigma$ (sigma), the mean becomes $\mu$ (mu), and the divisor is $n$ instead of $n - 1$. The word typical is loose on purpose: $s$ is the square root of an average squared distance, not the average of the distances.

Take the five values 4, 8, 11, 13, 14. The mean is $50/5 = 10$, so the deviations are -6, -2, 1, 3, and 4. Squared they are 36, 4, 1, 9, and 16, which sum to 66. Divide by $n - 1 = 4$ for the variance $s^2 = 16.5$, then take the square root: $s = 4.06$. Treating the same five numbers as a whole population instead gives $\sigma = \sqrt{66/5} = 3.63$, so settle that question before starting.

The sentence to unlearn is "the standard deviation is the average distance from the mean." For those five values the average distance really is $(6 + 2 + 1 + 3 + 4)/5 = 3.2$, which is the mean absolute deviation, and it is not 4.06. Squaring before averaging gives far-out values more weight, so unless every value sits the same distance out, $s$ lands above the plain average distance.

A standard deviation is never negative, and it equals 0 in exactly one case: every value in the set is identical, so every deviation is 0. It is also not resistant. Change that 14 to a 44 and the median stays at 11 while $s$ goes from 4.06 to 16.02, because one distance of 28 becomes 784 inside the sum and swamps the other four.

Report it with the variable and the units attached, never as a bare number. In the Fall 2026 course this sits in Unit 1, whose topic 1.7 is titled Summary Statistics for One Quantitative Variable.

Finance uses this exact statistic as its measure of risk. The standard deviation of an investment's period returns is computed by the same steps used here and is the number quoted as an asset's risk: standard deviation of returns.