Deviation

A deviation belongs to a single observation, not to a data set. It is written $x_i - \bar{x}$, where $x_i$ is one value and $\bar{x}$ (x-bar) is the mean of all of them, and it keeps the units of the data. The sign is part of the answer: positive puts the value above the mean, negative below it.

For 4, 7, 9, 12, 13 the mean is $45/5 = 9$, so the five deviations are -5, -2, 0, 3, and 4. They add to exactly 0, which is not a fact about this data set: $\sum (x_i - \bar{x}) = \sum x_i - n\bar{x}$, and $n\bar{x}$ is the total by definition, so the terms cancel every time. Squaring first rescues the idea. Those squared deviations are 25, 4, 0, 9, and 16, summing to a sum of squares of 54, which divided by $n - 1 = 4$ gives a variance of 13.5 and a standard deviation of 3.67.

"The deviation of this data set is 3.67" is a category error and the most common one on this page. Five values have five deviations; 3.67 is the standard deviation, one summary built from all of them. Calling one number the deviation of a whole set throws away the distinction that makes the formula work.

The exact-zero check only holds on an unrounded mean. On a set whose mean comes out as 9.333, rounding it to 9.3 before subtracting leaves the deviations missing zero by a little, a rounding artifact and not an error in the data. A deviation is also not a residual: a deviation measures a value against the mean of its own variable, while a residual, $y - \hat{y}$ (y minus y-hat), measures an observed response against what a model predicted.

Absolute values are the other way to stop the cancellation. Averaging $|x_i - \bar{x}|$ gives the mean absolute deviation, $(5 + 2 + 0 + 3 + 4)/5 = 2.8$ here. Squaring wins out because squared deviations have algebra behind them that absolute values do not, which is why $s$ carries through the rest of the course.