1.9 Mean Deviation
The obvious way to measure spread is to ask how far a typical observation lies from the mean. The deviations \(X-\overline {X}\) cannot simply be averaged, because they always sum to zero – the positive and negative ones cancel exactly. The mean deviation solves this by taking absolute values first, so that a deviation of \(-3\) counts as much as one of \(+3\).
Definition 1.35. \[\text {M.D}=\frac {\sum |X-\overline {X}|}{n}\quad \text {(ungrouped)}, \qquad \text {M.D}=\frac {\sum f|X-\overline {X}|}{\sum f}\quad \text {(grouped)}.\]
Solution. The mean is \(\overline {X}=\frac {28}{7}=4\). The absolute deviations are \[3,\ 2,\ 1,\ 0,\ 1,\ 2,\ 3,\qquad \sum |X-\overline {X}|=12,\] so \[\text {M.D}=\frac {12}{7}=1.71.\]
Solution. First the mean: \(\sum f=35\) and \(\sum fX=112\), so \(\overline {X}=\frac {112}{35}=3.2\).
| \(X\) | 1 | 2 | 3 | 4 | 5 | 6 | Total |
| \(f\) | 4 | 6 | 12 | 6 | 6 | 1 | 35 |
| \(|X-\overline {X}|\) | 2.2 | 1.2 | 0.2 | 0.8 | 1.8 | 2.8 | |
| \(f|X-\overline {X}|\) | 8.8 | 7.2 | 2.4 | 4.8 | 10.8 | 2.8 | 36.8 |
\[\text {M.D}=\frac {\sum f|X-\overline {X}|}{\sum f}=\frac {36.8}{35}=1.05.\]
Example 1.38. Find the mean deviation of the grouped data below.
| C-I | \(1\)–\(5\) | \(6\)–\(10\) | \(11\)–\(15\) | \(16\)–\(20\) |
| \(f\) | 1 | 3 | 4 | 2 |
Solution. Each class is represented by its mid-point. With \(\sum f=10\) and \(\sum fX=115\), the mean is \(\overline {X}=11.5\).
| C-I | Mid-point \(X\) | \(f\) | \(|X-\overline {X}|\) | \(f|X-\overline {X}|\) |
| \(1\)–\(5\) | 3 | 1 | 8.5 | 8.5 |
| \(6\)–\(10\) | 8 | 3 | 3.5 | 10.5 |
| \(11\)–\(15\) | 13 | 4 | 1.5 | 6.0 |
| \(16\)–\(20\) | 18 | 2 | 6.5 | 13.0 |
| Total | 10 | 38.0 |
\[\text {M.D}=\frac {38.0}{10}=3.8.\]
Remark 1.39. The mean deviation is easy to interpret – it is literally the average distance from the mean – but it is little used beyond an introductory course. The absolute value is awkward to handle algebraically, having no derivative at zero, so results about it are hard to obtain. Squaring instead of taking absolute values removes the signs just as effectively and leaves an expression that can be worked with, which is why the variance and standard deviation became the standard measures.
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