Mean absolute deviation for ungrouped data
Let $x_i, i=1,2, \cdots , n$ be $n$ observations.
The mean of $X$ is denoted by $\overline{x}$ and is given by
$$ \begin{eqnarray*} \overline{x}& =\frac{1}{n}\sum_{i=1}^{n}x_i \end{eqnarray*} $$
Formula:
$MAD =\dfrac{1}{n}\sum_{i=1}^{n}|x_i -\overline{x}|$
where,
$n$total number of observations$\overline{x}$sample mean
Example 1
The age (in years) of 6 randomly selected students from a class are : 22,25,24,23,24,20.
Compute mean absolute deviation about mean.
Solution
| $x_i$ | $(x_i-xb)$ | $|x_i-xb|$ | |
|---|---|---|---|
| 22 | -1 | 1 | |
| 25 | 2 | 2 | |
| 24 | 1 | 1 | |
| 23 | 0 | 0 | |
| 24 | 1 | 1 | |
| 20 | -3 | 3 | |
| Total | 138 | 8 |
The sample mean of $X$ is
$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{138}{6}\\ &=23 \text{ years} \end{aligned} $$
The mean absolute deviation about mean is
$$ \begin{aligned} MAD & =\dfrac{1}{n}\sum_{i=1}^{n}|x_i -\overline{x}|\\ &=\frac{8}{6}\\ &=\frac{8}{6}\\ &= 1.3333 \text{ years} \end{aligned} $$
Example 2
The systolic blood pressure (in mmHg) of 10 randomly selected patients are :
123, 128, 136, 112, 143, 114, 104, 137, 145, 150.
Compute mean absolute deviation about mean.
Solution
| $x_i$ | $(x_i-xb)$ | $|x_i-xb|$ | |
|---|---|---|---|
| 123 | -6.2 | 6.2 | |
| 128 | -1.2 | 1.2 | |
| 136 | 6.8 | 6.8 | |
| 112 | -17.2 | 17.2 | |
| 143 | 13.8 | 13.8 | |
| 114 | -15.2 | 15.2 | |
| 104 | -25.2 | 25.2 | |
| 137 | 7.8 | 7.8 | |
| 145 | 15.8 | 15.8 | |
| 150 | 20.8 | 20.8 | |
| Total | 1292 | 130 |
The sample mean of $X$ is
$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{1292}{10}\\ &=129.2 \text{ mmHg} \end{aligned} $$
The mean absolute deviation about mean is
$$ \begin{aligned} MAD & =\dfrac{1}{n}\sum_{i=1}^{n}|x_i -\overline{x}|\\ &=\frac{130}{10}\\ &=\frac{130}{10}\\ &= 13 \text{ mmHg} \end{aligned} $$
