MAD vs Standard Deviation

The mean absolute deviation answers one plain question: on average, how far does a value sit from the mean? Drop the sign on every deviation and average what is left, giving $\text{MAD} = \frac{1}{n}\sum |x_i - \bar{x}|$, where $\bar{x}$ (x-bar) is the mean and $n$ is the count. The divisor is $n$, not $n - 1$. There is no bias correction here, which is a real difference from the sample standard deviation.

For 2, 4, 6, 8 the mean is 5, the distances are 3, 1, 1, 3, and the MAD is $8/4 = 2$. Read that as: a typical value sits about 2 units from the mean. The sample standard deviation of the same four numbers is 2.58, larger, because squaring hands the two distances of 3 more say than the two distances of 1.

"MAD skips the squaring, so it is the outlier-resistant version of the standard deviation." It is not resistant at all. Change 1, 2, 3 into 1, 2, 1000 and the MAD goes from 0.67 to 443.78 while $s$ goes from 1 to 576.48. Both chase the outlier, and which one chases harder depends on how you ask. Here $s$ gains more in absolute terms, 575.48 against 443.11, while MAD gains more as a multiple, 665.7 times against 576.5 times. There is no general rule that MAD is the less inflated of the two. The resistant summaries of spread are the median and the IQR.

One inequality does hold for every data set: $\text{MAD} \le \sigma \le s$, where $\sigma$ (sigma) is the population standard deviation. MAD ties $\sigma$ only when every value sits the same distance from the mean, as with 10 and 20, where both come to 5. Otherwise MAD is strictly smaller.

Watch the abbreviation. Some software writes MAD for the median absolute deviation, the median of the distances from the median, which is a different number and a genuinely resistant one. In an introductory course MAD means the mean version about the mean.

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.