Measures of dispersion around the arithmetic mean, calculated as the average of absolute deviations from the mean or median.
## Core concept
Mean Deviation (MD) is the arithmetic average of the absolute differences between each data value and a measure of central tendency (usually the mean or median). It quantifies how far, on average, data points deviate from the centre.
Unlike variance or standard deviation, mean deviation uses absolute values rather than squares, making it less sensitive to extreme values. It is particularly useful when you need an intuitively interpretable measure of spread that is not dominated by outliers.
Key property: Mean deviation is always calculated from the arithmetic mean or median. When calculated from the mean, it is mathematically minimal compared to deviations from any other point.
## Formula / rule
For ungrouped data (from mean): $$\text{MD} = \frac{\sum|X_i - \bar{X}|}{n}$$
where: - $X_i$ = individual observations - $\bar{X}$ = arithmetic mean - $n$ = number of observations - $|\cdot|$ = absolute value (ignores sign)
For grouped data (from mean): $$\text{MD} = \frac{\sum f_i|X_i - \bar{X}|}{\sum f_i}$$
where $f_i$ = frequency of class $i$, $X_i$ = class midpoint
For ungrouped data (from median): $$\text{MD} = \frac{\sum|X_i - M|}{n}$$
where $M$ = median
Coefficient of Mean Deviation: $$\text{Coefficient of MD} = \frac{\text{MD}}{\text{Mean (or Median)}} \times 100$$
Used to compare dispersion across datasets with different units or magnitudes.
## Common exam applications
- Comparing variability: When datasets have different means, use coefficient of MD to standardise comparison.
2. Quality control: Manufacturing processes use MD to monitor consistency in product specifications.
3. Income distribution analysis: Preferred over SD when studying economic inequality, as extreme outliers don't distort the measure as heavily.
4. Identifying stability: Lower MD indicates more consistent data; higher MD suggests greater fluctuation.
## Worked example
Dataset: 4, 6, 8, 10, 12 (ungrouped data)
Step 1: Calculate mean $\bar{X} = \frac{4+6+8+10+12}{5} = \frac{40}{5} = 8$
Step 2: Find absolute deviations from mean - $|4 - 8| = 4$ - $|6 - 8| = 2$ - $|8 - 8| = 0$ - $|10 - 8| = 2$ - $|12 - 8| = 4$
Step 3: Sum and divide by n $$\text{MD} = \frac{4+2+0+2+4}{5} = \frac{12}{5} = 2.4$$
Step 4: Coefficient of MD $$\text{Coefficient} = \frac{2.4}{8} \times 100 = 30\%$$
## Common mistakes
- Forgetting absolute values: Using signed deviations defeats the purpose; always take absolute values.
- Confusing with standard deviation: MD and SD are different measures. MD uses absolute deviations; SD uses squared deviations and is therefore more sensitive to outliers.
- Wrong denominator in grouped data: Use $\sum f_i$, not total number of classes.
- Mixing mean and median: Decide upfront which central tendency to use; the calculation method differs slightly.
- Misinterpreting coefficient: Remember to multiply by 100 to express as a percentage; compare coefficients across datasets, not absolute MD values.
For exam practice, focus on calculation accuracy in both ungrouped and grouped data scenarios, and understand when MD is preferred over other dispersion measures (particularly when robustness to outliers matters).