How far is each value from the mean?
The range measures spread with only two values, the largest and the smallest. The mean absolute deviation uses every value. It asks how far each value is from the mean, and then takes the average of those distances.
Take the four values 1, 3, 6 and 10. Their mean is (1 + 3 + 6 + 10) ÷ 4 = 20 ÷ 4 = 5. The mean is often written x̄, read “x bar”, so here x̄ = 5.
The four values as bars, with the dashed mean line at 5. Two bars end below the line and two above it.
The differences with their signs add up to 0
Subtract the mean from each value: 1 − 5 = −4, 3 − 5 = −2, 6 − 5 = +1 and 10 − 5 = +5. A value below the mean gives a negative difference, and a value above it gives a positive one.
Now add them: −4 − 2 + 1 + 5 = 0. The differences below the mean cancel the differences above it exactly.
This happens for every data set, not only this one. The mean is the total divided by the number of values, so the total of n values is n × x̄. Subtracting x̄ from each of the n values subtracts n × x̄ altogether, which leaves 0. Here the total is 20, and 20 − 4 × 5 = 0. So the average of the signed differences is always 0, whether the values are packed together or spread far apart, and it cannot measure spread.
Σ(x − x̄) = 0 for every data set, so the signs come off before the average measures anything
Try to make the two sides of the balance different lengths.
The values 1, 3, 6 and 10, with their mean. The top bar sets the negative differences against the positive ones, and the two sides are always the same length. Drag the last value: the mean moves, the signed differences still add up to 0, and the bottom bar, the distances laid end to end, changes.
A distance carries no sign
The fix is to drop the signs. A value 4 below the mean is as far from it as a value 4 above it, so both are at a distance of 4. The distance of a value x from the mean is the size of x − x̄ without its sign, written |x − x̄| and called its absolute value: |−4| = 4 and |+5| = 5.
The distances of 1, 3, 6 and 10 from 5 are 4, 2, 1 and 5. None of them is negative, so they cannot cancel.
The differences from the mean total 0. The distances, the same numbers without their signs, total 12.
The mean of the distances
Now take the mean of the distances: (4 + 2 + 1 + 5) ÷ 4 = 12 ÷ 4 = 3. This is the mean absolute deviation, often shortened to MAD. In symbols, MAD : add up the distances from the mean and divide by the number of values.
The mean absolute deviation is in the same units as the data. It says that, on average, a value is 3 away from the mean. A larger MAD means the values are more spread out from their mean.
Same mean, packed tighter
The values 4, 5, 5 and 6 also have mean 5, since (4 + 5 + 5 + 6) ÷ 4 = 20 ÷ 4 = 5. Their distances from 5 are 1, 0, 0 and 1, so their mean absolute deviation is (1 + 0 + 0 + 1) ÷ 4 = 2 ÷ 4 = 0.5.
The two sets have the same mean, but the mean absolute deviations are 3 and 0.5. On average a value in the first set is 3 from the mean, and a value in the second set is only 0.5 from it. The second set is packed much more tightly around its mean.
The values 4, 5, 5 and 6 all end within 1 of the mean line at 5. Their mean absolute deviation is 0.5.
Working in a table
A runner times five laps of a track: 62, 65, 58, 70 and 60 seconds. Find the mean first, because every distance is measured from it: the total is 62 + 65 + 58 + 70 + 60 = 315 seconds, and 315 ÷ 5 = 63 seconds.
Then make a table with one row for each lap: the time, its difference from 63, and its distance from 63. The differences are −1, +2, −5, +7 and −3. They add up to 0, which is a check that the mean is right. The distances are 1, 2, 5, 7 and 3, and they add up to 18.
The mean absolute deviation is 18 ÷ 5 = 3.6 seconds. On average, a lap time is 3.6 seconds from the mean of 63 seconds. The range, 70 − 58 = 12 seconds, gives the full width of the times; the mean absolute deviation gives the typical distance of a time from the mean, and it uses all five laps.
The lap times, their differences from the mean of 63 seconds, and their distances from it. The distances total 18, so the mean absolute deviation is 18 ÷ 5 = 3.6 seconds.
Looking ahead: squaring instead
Dropping the signs is one way to stop the differences canceling. The other way is to square each difference, since a square is never negative. That is how the standard deviation, the next measure of spread, is built.