In a data set, absolute deviation of an element is the absolute difference between a given point and that element. Typically the point from which the deviation is measured is a measure of central tendency, most often we consider median or sometimes the mean of the data set. In this page, we shall learn in detail about absolute deviation.

## Definition

In statistical distribution, absolute deviation is an absolute differences between individual number and their median or mean.

Mainly we can find the absolute deviation for the mean and median:
1) Mean Absolute Deviation
2) Median Absolute Deviation
Mean Absolute Deviation
Mean deviation is defined as the average of the absolute deviations taken from an average usually, mean, median or mode. The mean absolute deviation is also calculated by subtracting each score from its median, taking the absolute value and then dividing by N (total number of scores).

Median Absolute Deviation

Median absolute deviation is defined as the median of the absolute deviation taken from median. The formula for median absolute deviation:

$MAD$ = | $y$i - Median of data set|
The mean absolute deviation is somewhat different than other measures of dispersion. This measure could be used with any measure of central tendency.

## Formula

Absolute deviation is the difference between particular number (mean/median) and a given point.
Absolute Deviation Equation:
$D$i = |$y$i - $m(y)$|
where: $D$i = absolute deviation,
$m(y)$ = measure of mean/median
and $y$i = data element

## Examples

Below are solved examples to illustrate absolute deviation:
Example 1: Find the absolute deviation for the data set {1, 2, 5, 6, 7, 8, 10, 15,13, 14, 11, 9, 10}.

Solution:

Step 1: Arrange given data in ascending order.

1, 2, 5, 6, 7, 8, 9, 10, 10, 11, 13, 14, 15

Step 2: Median $m(y)$ = 9
 $y$ $y$ - $m(y)$ |$y$ - $m(y)$| 1 -8 8 2 -7 7 5 -4 4 6 -3 3 7 -2 2 8 -1 1 9 0 0 10 1 1 10 1 1 11 2 2 13 4 4 14 5 5 15 6 6 $\sum$ = 111 $\sum$ = 44

Example 2: Calculate absolute deviation about mean for data set: {12, 4, 31, 23, 25}

Solution:

Let y = 12, 4, 31, 23, 25

Step 1 -
Mean of given data = $\frac{Sum\ of\ Observations}{Total\ number\ of\ Observations}$

= $\frac{12+4+ 31+ 23+ 25}{5}$

= $\frac{95}{5}$

= 19

=> Mean ($\bar{y}$) = 19

Step 2:
Absolute deviation about mean

 y y - $\bar{y}$ |y - $\bar{y}$| 12 -7 7 4 -15 15 31 12 12 23 4 4 25 6 6 $\sum$ = 95 $\sum$ = 44

## How to Calculate Absolute Deviation?

When the absolute value of the deviation is calculated, a different distribution emerges:
The following steps are involved in the calculation of absolute deviations:
1) Calculate the median or mean of a given series.
2) Subtract each score from the median or the mean.
3) Take the absolute value.
The absolute values of the scores from the mean, median and mode are shown below:

 $y$ $y$ - $\bar{y}$ |$y$ - $\bar{y}$| $y$ - $y$median |$y$ - $y$median| $y$ - $y$mode |$y$ - $y$mode| 6 1.714 1.714 2 2 3 3 5 0.714 0.714 1 1 2 2 7 2.714 2.714 3 3 4 4 3 -1.286 1.286 -1 1 0 0 3 -1.286 1.286 -1 1 0 0 4 -0.286 0.286 0 0 1 1 2 -2.286 2.286 -2 2 -1 1 $\sum$ = 30 $\sum$ = 10.286 $\sum$ = 10 $\sum$ = 11

From the above table we can see that, the total of the absolute values of the median is the smallest value.

## Absolute Standard Deviation

Standard deviation is the most reliable and most commonly used absolute measure of dispersion. The standard deviation method is derived from the mean absolute deviation. The basic difference between standard deviation and the mean absolute deviation is that in the latter each deviation from the mean is squared.
The absolute standard deviation is the square root of the sum of the squared deviance of all values from the absolute mean, divided by the number of values.The equation for ungrouped data from a sample is a variation of that for the mean deviation:

Standard deviation = $\sqrt{\sum|y-\bar{y}|^2}{n}$
Where, $\bar{y}$ = mean of the data and
$n$ = total number of scores.

## Calculate Average Deviation

Average deviation is a condensed statistic of statistical dispersion. We can calculate average deviation by taking the average of the absolute deviations. In general form, average is the result of measure of central tendency.
Average deviation is help to summarize a set of observations, variability or statistic of statistical dispersion. It is the average of the absolute deviations and also called the mean absolute deviation.

The average deviation for the set {$y_1, y_2, y_3,....., y_k$} is given below:

$\frac{1}{k}$ $\sum_{i=1}^k$ |$y$i - $m(y)$|

$n$ = Total score
$m(y)$ = Measure of central tendency
Follow the following steps to find the average absolute deviation:

Step 1: Subtract the mean from each score in order to get the distance of that score form the mean.

Step 2: Consider all these deviations from the mean as positive.

Step 3:
Take the average of these absolute deviation.
Example: Find the average deviation for the data set {2, 4, 6, 2, 3} and show that total of the absolute values of the median is the smallest value.

Solution:

Arrange the data in ascending order.
{2, 2, 3, 4, 6}
Total number of terms (n) = 5

Mean = $\frac{2+2+3+4+6}{5}$ = 3.4
Median = 3
Mode = 2

 $y$ |$y$ - $\bar{y}$| |$y$ - $y$median| |$y$ - $y$mode| 2 1.4 1 0 2 1.4 1 0 3 0.4 0 1 4 0.6 1 2 6 2.6 3 4 $\sum$ = 6.4 $\sum$ = 6 $\sum$ = 7

Now
$\frac{1}{n}$$\sum |y - \bar{y}| = \frac{1}{5} \times 6.4 = 1.28 \frac{1}{n} \sum|y - ymedian| = \frac{1}{5} \times 6 = 1.2 (smallest value) \frac{1}{n}$$\sum$ |$y$ - $y$mode| = $\frac{1}{5}$ $\times$ 7 = 1.4