Mean absolute deviation (MAD): formula and examples

What the mean absolute deviation measures

The mean absolute deviation answers one plain question: on average, how far is a value from the mean? The formula says exactly that.

Here $\bar{x}$ ("x-bar") is the sample mean, $n$ is how many values you have, and $|x_i - \bar{x}|$ is the distance from one value to the mean with the sign thrown away. The vertical bars are absolute value, so a value 3 below the mean and a value 3 above the mean both contribute 3.

The absolute value is doing necessary work. Deviations from the mean always add to exactly zero, because the mean sits at the balance point of the data. For 2, 4, 6, 8 the mean is 5 and the deviations are -3, -1, 1, and 3, which sum to 0. Averaging raw deviations would report zero spread for every data set on earth. Absolute value fixes that by making every deviation count as a distance.

MAD comes out in the units of the data. For commute times in minutes, a MAD of 2.3 means "the typical commute sits about 2.3 minutes from the average commute." That plain reading is MAD's main selling point over the standard deviation, which needs a squaring step and a square root before it lands back in minutes.

How to find MAD in four steps

1. Find the mean $\bar{x}$.
2. Subtract the mean from every value to get the deviations $x_i - \bar{x}$.
3. Drop the sign on each one, giving $|x_i - \bar{x}|$.
4. Add those distances and divide by $n$.

Step 4 divides by $n$, not by $n - 1$. That is a real difference from the sample standard deviation, which uses $n - 1$ to correct a bias in estimating a population variance. MAD has no such correction in standard use, so it always divides by the count. If you are unsure why the standard deviation is different, see why divide by n minus 1.

One warning about the abbreviation. Some software and some statisticians use MAD for the median absolute deviation, the median of the distances from the median, which is a different and much more outlier-resistant number. In an introductory course MAD means mean absolute deviation about the mean, but check the definition before you trust a printout.

MAD is never larger than the standard deviation

For any data set at all, the mean absolute deviation is less than or equal to the population standard deviation, and the population standard deviation is less than or equal to the sample standard deviation:

So MAD is at most as large as either version of the standard deviation. Four data sets, all computed the same way:

The last row is the one case where MAD ties the population standard deviation. With only the two values 10 and 20, both deviations from the mean of 15 have size 5, so squaring and unsquaring changes nothing. That is the general rule: MAD equals $\sigma$ exactly when every value sits the same distance from the mean, and MAD is strictly smaller otherwise.

The reason is short, and it also delivers the equality case. Write $d_i = x_i - \bar{x}$ for the deviations. Then $\sigma^2$ is the average of the squared distances $d_i^2$, while MAD is the average of the distances $|d_i|$ themselves. So $\sigma^2 - \text{MAD}^2$ is the average of the squared distances minus the square of their average, which is exactly the variance of those distances, and no variance is ever negative. That forces $\sigma^2 \ge \text{MAD}^2$, with equality only when the distances have no spread at all, meaning every value sits the same distance from the mean. Squaring weights the far-out distances more heavily than the near-in ones, so the more the distances vary among themselves, the wider the gap: in the third row, one value at 78 makes $\sigma$ about 1.43 times MAD, while in the first row the ratio is only about 1.12.

For a normal population there is a fixed ratio. The mean absolute deviation equals $\sigma\sqrt{2/\pi}$, which is about $0.798\sigma$. That is one reason MAD never took over: on the normal curve the course is built around, it carries the same information as $\sigma$ but does not plug into anything.

Why statistics squares instead of taking absolute values

MAD is easier to explain, so why is the standard deviation everywhere instead? Four reasons, in rough order of how much they matter.

Squares are smooth and absolute values are not. The function $f(d) = d^2$ has a derivative everywhere. The function $f(d) = |d|$ has a corner at $d = 0$ where no derivative exists. Almost every result in statistics is derived by setting a derivative to zero and solving, and that whole machine stalls on a corner. Least-squares regression is the clearest case: you can find the least-squares regression line in closed form precisely because the thing being minimized is a smooth sum of squares. The absolute-value version has no such tidy formula.

Squared deviations point at the mean; absolute deviations point at the median. The mean is the number that makes $\sum (x_i - c)^2$ as small as possible. The median is a number that makes $\sum |x_i - c|$ as small as possible; with an even count, so is anything between the two middle values. Check it on 7, 8, 8, 9, 10, 12, 13, 15, where the mean is 10.25 and the median is 9.5: the sum of absolute deviations is 18.5 about the mean but only 18 about the median, while the sum of squared deviations is 55.5 about the mean and 60 about the median. Each center has its own natural measure of spread, so pairing the mean with an absolute-value spread is a slightly mismatched combination.

Variances add and MADs do not. For independent random quantities, variances combine by adding, which is what makes standard errors work. There is no comparable rule for MAD, so it cannot carry you into inference.

The standard deviation is already load-bearing. It defines the z-score, it drives the empirical rule, and it sits inside every standard error and margin of error later in the course. Replacing it would mean rebuilding all of that. For more on the squaring step itself, see why is variance in squared units and why square the residuals.

What an outlier does to MAD and to the standard deviation

Neither measure is resistant. Both use the mean, and both use every value, so a single extreme observation moves both. But it does not move them equally.

Take seven commute times: 12, 14, 15, 15, 16, 18, 22 minutes. The mean is 16, MAD is 2.2857, and the sample standard deviation is 3.2146. Now change the last value to 78 minutes, as though one trip hit a road closure. The mean jumps to 24, MAD jumps to 15.4286, and $s$ jumps to 23.8817.

MAD grew by a factor of about 6.75. The standard deviation grew by a factor of about 7.43. The squaring step gave the one enormous deviation extra leverage, so the standard deviation reacted harder. Be careful how you state that comparison, because it is a claim you often see put too strongly. The safe form is the absolute one: an outlier moves $s$ further than it moves MAD, so the gap between the two widens. Comparing growth factors, as the numbers above do, is a different question, and with only a handful of values MAD is sometimes the one that multiplies by more. Neither version makes MAD resistant. If you want a genuinely resistant measure of spread, use the interquartile range instead. See what is a resistant statistic and standard deviation vs IQR.

Where MAD fits the AP course

Topic 1.7 of the Fall 2026 AP Statistics course names three measures of variability: the range, the interquartile range, and the standard deviation. The mean absolute deviation is not one of them, and it is not on the AP formula sheet, so you will not be asked to compute a MAD on the exam.

It is still worth the ten minutes. Middle school and pre-AP courses teach MAD first because the idea of "average distance from the mean" is exactly what a standard deviation is trying to be, minus the squaring. If you can explain MAD, you can explain what $s$ is measuring, which is the interpretation the AP exam does ask for: a typical distance between the values and their mean. Many introductory college courses also use MAD as a stepping stone for the same reason.

When a question asks you to describe spread, report an IQR or a standard deviation with units. Which of the two you should report is decided by the shape of the distribution, covered in how to describe the shape of a distribution. To compute the standard deviation itself, follow standard deviation by hand or use the standard deviation calculator.