Range and mean deviation

The range of a dataset is the simplest measure of dispersion: largest value minus smallest. It uses only two numbers, so it ignores everything in between. The mean deviation is a much better measure: the average of all deviations from the centre.

Range

For data $x_1, \dots, x_n$: $\text{Range} = \max(x_i) - \min(x_i).$

For grouped data, take the upper limit of the highest class minus the lower limit of the lowest class.

Range is quick to compute but unduly influenced by outliers and uses no information about the bulk of the data. We need finer measures.

Mean deviation about a centre

Pick a centre $C$ (typically mean $\bar{x}$ or median $M$). The mean deviation about $C$ is the average of absolute deviations: $\text{MD}_C = \frac{\sum_{i=1}^n |x_i - C|}{n}.$

For frequency data: $\text{MD}_C = \frac{\sum f_i |x_i - C|}{N}, \quad N = \sum f_i.$

For grouped data, use class midpoints for $x_i$.

The two most common choices are $\bar{x}$ (mean) and $M$ (median). It turns out that mean deviation is minimised when the centre is the median , that is, $\text{MD}_M \le \text{MD}_C$ for any $C$.

Steps to compute

1. Find the centre ($\bar{x}$ or $M$).
2. Compute $|x_i - C|$ for each $i$.
3. Multiply by $f_i$ if frequency data.
4. Sum and divide by $n$ (or $N$).

Worked examples

Example 1. Mean deviation about the mean of $2, 4, 6, 8, 10$.

$\bar{x} = 6$. Deviations: $|2 - 6| = 4, 2, 0, 2, 4$. Sum = $12$. MD = $12/5 = 2.4$.

Example 2. Mean deviation about the median of $5, 8, 12, 15, 20, 22, 25$.

$n = 7$, median = $x_4 = 15$. Deviations: $10, 7, 3, 0, 5, 7, 10$. Sum = $42$. MD = $42/7 = 6$.

Example 3. Mean deviation about the mean of frequency data:

$N = 10$. $\sum f x = 2 + 12 + 24 + 16 = 54$. $\bar{x} = 5.4$.

$|x_i - 5.4|$: $3.4, 1.4, 0.6, 2.6$.

$f_i |x_i - 5.4|$: $3.4, 4.2, 2.4, 5.2$. Sum = $15.2$. MD = $15.2/10 = 1.52$.

Example 4. Range of $12, 8, 15, 21, 7, 19$.

$\max = 21, \min = 7$. Range $= 14$.

Example 5. For grouped data:

Midpoints: $5, 15, 25, 35$. $N = 35$. $\sum f x = 25 + 120 + 375 + 245 = 765$. $\bar{x} = 765/35 = 21.857$.

$|x - \bar{x}|$: $16.857, 6.857, 3.143, 13.143$. $f |x - \bar{x}|$: $84.286, 54.857, 47.143, 92.0$. Sum $\approx 278.286$. MD $\approx 278.286/35 \approx 7.95$.

Try it yourself

1. Range of $3, 7, 11, 4, 9, 15$.
2. Mean deviation about mean of $10, 20, 30, 40, 50$.
3. Mean deviation about median of $4, 6, 9, 12, 15$.
4. For data $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, find the mean deviation about the mean.
5. Compute mean deviation about mean of $5, 5, 5, 5, 5$. (What do you expect?)
6. Find mean deviation about the median of $6, 11, 14, 22, 28$.
7. For frequency data with values $1, 2, 3, 4$ and frequencies $4, 5, 3, 2$, find the MD about the mean.
8. Show that for any constant $c$, MD about mean $\le$ MD about $c$. Or show MD about median $\le$ MD about $c$.
9. Find MD about median of $5, 10, 15, 20, 25, 30$.
10. Range and mean deviation about median of $2, 4, 6, 6, 8$.
11. For data $1, 2, 4, 5, 6, 8, 10$, compute both the mean deviation about the mean and median. Which is smaller?
12. The marks of $10$ students are $20, 22, 25, 25, 26, 28, 30, 33, 35, 40$. Find the mean deviation about the mean.

Pitfalls / Tricks

• Always take absolute value in the deviation.
• For grouped data, use midpoints , don't forget.
• Mean deviation about the median is the smallest among all choices of centre.
• Insight. Mean deviation is "average distance from the centre". It is intuitive but harder to manipulate algebraically , which is why variance (squared deviations) is preferred for theory.