Range, Interquartile Range and Mean Absolute Deviation

Why the center is only half the story

These two data sets have exactly the same mean and the same median.

$$A: \; 49, \; 50, \; 51 \qquad B: \; 10, \; 50, \; 90$$

Both have mean $50$ and median $50$. Reported that way they are identical, and they are not remotely alike.

A measure of spread is the number that separates them. It answers: how far from the center does the data actually sit?

This matters outside of textbooks. Two bus routes averaging $30$ minutes are not equally useful if one always takes $28$ to $32$ and the other varies between $10$ and $50$. The average is the same; only one is worth planning around.

Range: the full width

$$\text{range} = \text{largest} - \text{smallest}$$ $$A: 51 - 49 = 2 \qquad B: 90 - 10 = 80$$

The range separates those two sets instantly, and it is the quickest measure to compute.

Its weakness is that it uses only two values and throws the rest away. One outlier sets the range on its own:

$$5, 6, 6, 7, 7, 8, 94 \quad\Rightarrow\quad \text{range} = 89$$

Six of those seven values sit within $3$ of each other, and the range says $89$.

Interquartile range: the middle half

The quartiles cut sorted data into four parts. $Q_1$ is the median of the lower half and $Q_3$ is the median of the upper half.

$$\text{IQR} = Q_3 - Q_1$$

Take $4, 7, 9, 12, 14, 15, 21$. The median is $12$. With an odd count, leave the median out of both halves:

$$\text{lower } (4, 7, 9) \Rightarrow Q_1 = 7 \qquad \text{upper } (14, 15, 21) \Rightarrow Q_3 = 15$$ $$\text{IQR} = 15 - 7 = 8$$

The IQR is the width of the middle 50% of the data. Because it discards the top and bottom quarters, an outlier sitting out in one of those tails has no effect on it at all.

Data	Range	IQR
$5, 6, 6, 7, 7, 8, 9$	$4$	$2$
$5, 6, 6, 7, 7, 8, 94$	$89$	$2$

Replacing $9$ with $94$ multiplies the range by more than twenty and leaves the IQR untouched. That is exactly the behaviour you want when one value is unusual.

Mean absolute deviation: the average distance from the mean

Range and IQR each use two values. Mean absolute deviation (MAD) uses every one.

Three steps:

1. Find the mean.
2. Find how far each value is from it, ignoring the sign.
3. Average those distances.

For $3, 5, 8, 12$:

$$\text{mean} = \frac{28}{4} = 7$$

Value	Distance from $7$
$3$	$4$
$5$	$2$
$8$	$1$
$12$	$5$

$$\text{MAD} = \frac{4 + 2 + 1 + 5}{4} = \frac{12}{4} = 3$$

On average, values sit $3$ away from the mean.

The signs are dropped for a reason worth knowing. Distances above the mean and below it always cancel exactly — that is what being the mean means — so without dropping the signs the total would be $0$ for every data set ever collected, and would say nothing. Taking each distance as an absolute value is what keeps the measure alive.

Choosing between them

Measure	Uses	Affected by outliers?
range	$2$ values	heavily
IQR	$2$ quartiles	no
MAD	every value	yes, but spread out

Match the measure of spread to the measure of center:

• Reporting the mean? Report the MAD. Both use every value.
• Reporting the median? Report the IQR. Both resist outliers.

That pairing is the point. A median with a range attached mixes a measure that ignores extremes with one that is decided by them.

Worked examples

Common mistakes

Practice problems