Calculating the Mean Absolute Deviation | Grade 7 Math | LevelKite

Here are two small data sets.

Set A: $9, 10, 10, 11$
Set B: $4, 8, 12, 16$

Each set has $4$ values that add to $40$ , so each mean is $40 \div 4 = 10$ . The mean alone cannot tell these sets apart. Still, they differ: the values in Set A stay close to $10$ , and the values in Set B lie much farther from it.

Variability is how spread out the values of a data set are. You already know one way to measure it, the range. In this lesson you'll calculate another measure, the mean absolute deviation. It uses every value in the set, and it answers one question: on average, how far is a value from the mean?

Three ideas make "how far" exact.

Absolute value. The absolute value of a number $a$ , written $|a|$ , is the distance from $a$ to $0$ on the number line. A distance is never negative, so $|-6| = 6$ , $|6| = 6$ and $|0| = 0$ . The distance between any two numbers $p$ and $q$ is $|p - q|$ .

Order. A number is less than another number when it lies to the left of it on the number line. Order follows position on the line, not distance from $0$ : $-8$ is less than $-2$ , even though $|-8| = 8$ is greater than $|-2| = 2$ .

Deviation. Let $x$ be a data value and let $m$ be the mean of its data set. The deviation of $x$ is $x - m$ . It is negative when $x$ is less than $m$ , zero when $x$ equals $m$ , and positive when $x$ is greater than $m$ . The absolute deviation of $x$ is $|x - m|$ , the distance between $x$ and the mean. It is never negative.

The table shows both for each value of Set B, whose mean is $10$ .

Set B: each value compared with the mean, 10

Value	Deviation (value minus mean)	Absolute deviation (distance from mean)
$4$	$4 - 10 = -6$	$\lvert -6 \rvert = 6$
$8$	$8 - 10 = -2$	$\lvert -2 \rvert = 2$
$12$	$12 - 10 = 2$	$\lvert 2 \rvert = 2$
$16$	$16 - 10 = 6$	$\lvert 6 \rvert = 6$

So average the distances. For Set B, the absolute deviations are $6$ , $2$ , $2$ and $6$ . Add them, then divide the sum by the number of values, $4$ .

\frac{6 + 2 + 2 + 6}{4} = \frac{16}{4} = 4

The result, $4$ , is the mean absolute deviation of Set B. On average, a value in Set B is a distance of $4$ from the mean.

Set A also has mean $10$ . Its absolute deviations are $|9 - 10| = |-1| = 1$ , $|10 - 10| = 0$ , $|10 - 10| = 0$ and $|11 - 10| = 1$ . The value $10$ appears twice, so its distance, $0$ , counts twice. The sum is $1 + 0 + 0 + 1 = 2$ , and $2 \div 4 = 0.5$ . The mean absolute deviation of Set A is only $0.5$ .

The results $4$ and $0.5$ measure what the mean could not show: on average, the values of Set B lie farther from their mean than the values of Set A do.

Key concept

Mean absolute deviation (MAD)

The mean absolute deviation (MAD) of a data set is the mean of the distances between each value and the mean of the set.

Let $n$ be the number of values, where $n$ is at least $1$ . Call the values $x_1, x_2, \ldots, x_n$ and their mean $m$ .

The distance between two numbers $p$ and $q$ is $|p - q|$ , so the distance between a value $x$ and the mean is $|x - m|$ .
$|a|$ is the distance from $a$ to $0$ , so it is never negative. The MAD is never negative, and it is $0$ exactly when all the values are equal.
Order follows position on the number line, not distance from $0$ : a value left of $m$ is less than $m$ , and a value right of $m$ is greater. Either way, only its distance counts.
The MAD has the same unit as the data.

\text{MAD} = \frac{|x_1 - m| + |x_2 - m| + \cdots + |x_n - m|}{n}

You can calculate the MAD of any set of numbers that has at least one value. Follow four steps.

Find the mean, $m$ . Keep its exact value, because every distance is measured from it.
Find each absolute deviation, $|x - m|$ . Subtract first, then take the absolute value, because the distance between $x$ and $m$ is $|x - m|$ . A value that appears more than once gets a distance each time.
Add the absolute deviations.
Divide the sum by the number of values, $n$ .

Two checks help in step 2. First, a value left of $m$ on the number line must have a negative deviation, and a value right of $m$ must have a positive one. Second, with the exact mean, the deviations $x - m$ always add to $0$ before you take absolute values. If yours add to any other number, look for an arithmetic mistake.

Worked example

Whole-number data

Find the mean absolute deviation of the data set $4, 7, 9, 10, 10$ .

Find the mean, m, because every distance is measured from it. The set has 5 values, so add them and divide the sum by 5.
$4 + 7 + 9 + 10 + 10 = 40,\qquad m = 40 \div 5 = 8$
Subtract the mean, 8, from each value to find its deviation. The values 4 and 7 lie left of 8 on the number line, so they are less than the mean and their deviations are negative. The value 10 appears twice, so it gets two deviations.
$4 - 8 = -4,\quad 7 - 8 = -1,\quad 9 - 8 = 1,\quad 10 - 8 = 2,\quad 10 - 8 = 2$
Check the deviations. The mean is exact, so they must add to 0, and they do.
$(-4) + (-1) + 1 + 2 + 2 = -5 + 5 = 0$
Take the absolute value of each deviation. An absolute value is a distance from 0, so it is never negative. Each result is the distance between a value and the mean.
$|-4| = 4,\quad |-1| = 1,\quad |1| = 1,\quad |2| = 2,\quad |2| = 2$
Add the five distances to find the total distance from the mean.
$4 + 1 + 1 + 2 + 2 = 10$
Divide the total by the number of values, 5. Dividing shares the total distance equally among the values, which is what a mean does.
$\text{MAD} = 10 \div 5 = 2$

Answer

The mean absolute deviation of the data set is $2$ . On average, a value in the set is a distance of $2$ from the mean, $8$ .

Worked example

Temperatures below zero

The low temperatures on five winter days, in degrees Celsius, were $-8, -5, -1, 1, 3$ . Find the mean absolute deviation of these temperatures.

Find the mean, m. Add the three negative values, add the two positive values, then combine the two sums. Divide the result by 5, the number of values. The mean is −2 degrees Celsius.
$(-8) + (-5) + (-1) = -14,\quad 1 + 3 = 4,\quad -14 + 4 = -10,\quad m = -10 \div 5 = -2$
Find the deviations of −8 and −5 first. Order follows position on the number line, not distance from 0: both lie left of −2, so both are less than the mean and their deviations are negative. Subtracting −2 is the same as adding its opposite, 2.
$-8 - (-2) = -8 + 2 = -6,\qquad -5 - (-2) = -5 + 2 = -3$
Find the deviations of −1, 1 and 3. All three lie right of −2 on the number line, so they are greater than the mean and their deviations are positive. That includes −1, which is greater than −2 even though it is negative.
$-1 - (-2) = -1 + 2 = 1,\qquad 1 - (-2) = 1 + 2 = 3,\qquad 3 - (-2) = 3 + 2 = 5$
Check the deviations. The mean is exact, so they must add to 0, and they do.
$(-6) + (-3) + 1 + 3 + 5 = -9 + 9 = 0$
Take the absolute value of each deviation. Each result is the distance, in degrees Celsius, between a temperature and the mean. A distance is never negative.
$|-6| = 6,\quad |-3| = 3,\quad |1| = 1,\quad |3| = 3,\quad |5| = 5$
Add the five distances, then divide the sum by the number of values, 5, to find the mean distance.
$\text{MAD} = \frac{6 + 3 + 1 + 3 + 5}{5} = \frac{18}{5} = 3.6$

Answer

The mean absolute deviation is $3.6$ degrees Celsius. On average, a day's low temperature was $3.6$ degrees Celsius away from the mean of $-2$ degrees Celsius.

Worked example

A mean that is not a whole number

Six runners finished a race in $23$ , $28$ , $30$ , $32$ , $36$ and $40$ minutes. Find the mean absolute deviation of their times.

Find the mean, m, of the six times. Keep the exact value, 31.5 minutes, because every distance is measured from it. Rounding it to 32 would change every distance in this data set.
$23 + 28 + 30 + 32 + 36 + 40 = 189,\qquad m = 189 \div 6 = 31.5$
The distance between two numbers is the absolute value of their difference. Start with the three times that lie left of 31.5 on the number line. They are less than the mean, so their deviations are negative. Taking the absolute value gives each distance, which is never negative.
$|23 - 31.5| = |-8.5| = 8.5,\quad |28 - 31.5| = |-3.5| = 3.5,\quad |30 - 31.5| = |-1.5| = 1.5$
Do the same for the three times that lie right of 31.5. Their deviations are positive, so each one already equals its distance from 0, and taking the absolute value leaves it unchanged.
$|32 - 31.5| = |0.5| = 0.5,\quad |36 - 31.5| = |4.5| = 4.5,\quad |40 - 31.5| = |8.5| = 8.5$
Add the distances. The three below the mean total 13.5, and the three above the mean total 13.5. These two totals match whenever the mean is exact, because the deviations then add to 0.
$8.5 + 3.5 + 1.5 = 13.5,\qquad 0.5 + 4.5 + 8.5 = 13.5,\qquad 13.5 + 13.5 = 27$
Divide the total distance by the number of values, 6, to find the mean distance.
$\text{MAD} = 27 \div 6 = 4.5$

Answer

The mean absolute deviation is $4.5$ minutes. On average, a runner's time was $4.5$ minutes away from the mean time of $31.5$ minutes.

The mean absolute deviation describes variability with one number: the average distance between the values of a data set and their mean. To calculate it, find the mean, find each distance $|x - m|$ , add the distances and divide by the number of values. A distance is never negative, so nothing cancels, and a greater MAD means the values lie, on average, farther from the mean.

Next, you'll find a missing data value when you know the mean, median, mode or range. You'll also identify outliers, which are values far from the rest of a data set, and describe how they affect measures of center.