Finding the Mean Absolute Deviation | Grade 6 Math | LevelKite

Ana and Ben each played 4 rounds of a quiz game. Ana scored $5, 6, 6, 7$ points. Ben scored $3, 5, 6, 10$ points. Each list of scores is a data set.

Start with each player's mean. To find a mean, add the values, then divide by the number of values.

Ana: $5 + 6 + 6 + 7 = 24$ and $24 \div 4 = 6$ points.
Ben: $3 + 5 + 6 + 10 = 24$ and $24 \div 4 = 6$ points.

The means match, but the data sets do not look alike. Ana's scores stay close to $6$ . Ben's scores reach farther from $6$ . Variability is how spread out the values in a data set are.

You already know one measure of variability: the interquartile range (IQR), which is the third quartile minus the first quartile, $Q_3 - Q_1$ . This lesson gives you a second measure. It starts from the mean and uses every value in the data set.

To measure spread around the mean, ask one question about each value: how far is it from the mean?

That distance is the value's absolute deviation. Deviation means how far a value strays from the mean. Absolute means you keep only the distance, which is never negative. It does not matter whether the value is above or below the mean.

Picture the scores on a number line. The distance from a score to the mean is the gap between the two numbers. To find it, subtract the smaller number from the larger one. Ben's mean is $6$ points. His score of $10$ is $10 - 6 = 4$ points from the mean. His score of $3$ is $6 - 3 = 3$ points from the mean. A value equal to the mean is $0$ points away.

The table lists all of Ben's distances. Each row is one of his 4 scores. The first column is the score, in points. The second column is the subtraction. The third column is the distance from the mean, in points.

Ben's 4 scores and their distances from the mean of 6 points

Ben's score (points)	Subtraction	Distance from the mean (points)
$3$	$6 - 3$	$3$
$5$	$6 - 5$	$1$
$6$	$6 - 6$	$0$
$10$	$10 - 6$	$4$

Now turn the 4 distances into one number by finding their mean. Add the distances, then divide by 4, because there are 4 distances, one for each score. The distance of $0$ still counts as one of the 4.

This number is the mean absolute deviation, or MAD. The name says what it is: the mean of the absolute deviations. Notice that you find a mean twice: first the mean of the scores, then the mean of the distances.

\text{MAD} = \dfrac{3 + 1 + 0 + 4}{4} = \dfrac{8}{4} = 2 \text{ points}

Ben's MAD is $2$ points. On average, his scores are $2$ points away from his mean of $6$ points. The distances are in points, so the MAD is in points too.

Key concept

Mean absolute deviation (MAD)

The mean absolute deviation (MAD) of a data set of numbers is the mean of the distances between each data value and the mean of the data. It tells you how far the values are from the mean, on average.

To find it:

Find the mean of the data.
Find each value's distance from the mean. Subtract the smaller number from the larger one. A value equal to the mean has a distance of $0$ .
Add all the distances. Include every $0$ .
Divide the sum by the number of data values.

The MAD has the same unit as the data. When two data sets use the same unit, the one with the larger MAD has values that are farther from its mean, on average.

\text{MAD} = \dfrac{\text{sum of the distances from the mean}}{\text{number of data values}}

Worked example

Ana's scores

Ana scored $5, 6, 6, 7$ points in her 4 rounds. Find the mean absolute deviation (MAD) of her scores.

Find the mean first, because every distance is measured from it. Add the 4 scores, then divide by 4.
$5 + 6 + 6 + 7 = 24, \quad 24 \div 4 = 6$
Find each score's distance from the mean of 6 points. Subtract the smaller number from the larger one, so no distance is negative. Each score of 6 equals the mean, so its distance is 0.
$6 - 5 = 1, \quad 6 - 6 = 0, \quad 6 - 6 = 0, \quad 7 - 6 = 1$
Add the 4 distances. Keep both zeros, because every score has its own distance.
$1 + 0 + 0 + 1 = 2$
Divide the sum by 4, the number of scores. This gives the mean of the distances, which is the MAD.
$2 \div 4 = 0.5$

Answer

The MAD of Ana's scores is $0.5$ points. That is less than Ben's MAD of $2$ points. So Ana's scores are closer to the mean of $6$ points, on average, than Ben's scores are.

Worked example

When the mean is a decimal

A basketball player scored $4, 6, 7, 7, 9, 12$ points in 6 games. Find the mean absolute deviation (MAD) of the player's points per game.

Find the mean. Add the 6 values, then divide by 6. The mean is a decimal here. That is fine, because the mean does not have to be one of the data values.
$4 + 6 + 7 + 7 + 9 + 12 = 45, \quad 45 \div 6 = 7.5$
Four values are below the mean of 7.5 points. Subtract each one from 7.5 to find its distance. The value 7 appears twice, so write its distance twice.
$7.5 - 4 = 3.5, \quad 7.5 - 6 = 1.5, \quad 7.5 - 7 = 0.5, \quad 7.5 - 7 = 0.5$
Two values are above the mean. Subtract 7.5 from each one to find its distance.
$9 - 7.5 = 1.5, \quad 12 - 7.5 = 4.5$
Check before you go on. The mean of 7.5 is exact, so the total distance below it must equal the total distance above it. Both totals are 6, so the distances pass this check. If the totals were different, you would recheck the mean and each distance.
$3.5 + 1.5 + 0.5 + 0.5 = 6, \quad 1.5 + 4.5 = 6$
Add all 6 distances. The two totals from the check cover every distance, so add them.
$3.5 + 1.5 + 0.5 + 0.5 + 1.5 + 4.5 = 6 + 6 = 12$
Divide the sum by 6, the number of values. Then check with multiplication, which undoes division: the MAD times 6 gives back the sum of 12.
$12 \div 6 = 2, \quad 2 \times 6 = 12$

Answer

The MAD is $2$ points. On average, the player's points in a game are $2$ points away from the mean of $7.5$ points.

Takeaway: The mean absolute deviation (MAD) is the mean of the distances between the data values and their mean. It tells you how far the values are from the mean, on average. To find it, find the mean, find every value's distance from the mean, add the distances, and divide by the number of values.

Next, you will learn to identify an outlier, a value far from the rest of the data, and describe its effect. You will also practice identifying mistakes in finding measures of variability.