Throw away the extreme quarter at each end, and measure what is left.
Test for an outlier
A data set has Q1 = 18 and Q3 = 30. Which of these values would be flagged as an outlier?
IQR = 30 − 18 = 12, and 1.5 × 12 = 18. The fences sit at 18 − 18 = 0 and 30 + 18 = 48, so only a value below 0 or above 48 is an outlier — and that is 52. A is the trap: 2 looks far from the box, but it is still above the lower fence of 0. C at 46 is close to the fence and still inside it. Being unusual-looking is not the test; crossing a fence is.
IQR = Q3 − Q1
Q1 = 18, Q3 = 30
the quartiles
IQR = 30 − 18 = 12
the box's width
| Measure | What it uses | Upset by an extreme value? |
|---|---|---|
| Range | max − min | completely |
| IQR | Q3 − Q1 | no |
The range is decided entirely by the two most unusual values in the data. The IQR ignores the top and bottom quarters altogether, which is exactly why it is trusted.
lower fence = Q1 − 1.5 × IQR · upper fence = Q3 + 1.5 × IQR
An outlier is a value that lies far from the rest of the data. The fences turn "far" into a test: a value below the lower fence or above the upper fence is an outlier.
1.5 × 12 = 18
the fence distance
lower = 18 − 18 = 0
below this is an outlier
upper = 30 + 18 = 48
above this is an outlier
The distance added is 1.5 × IQR, measured outwards from each quartile. Multiplying Q1 or Q3 itself by 1.5 is a different number entirely.
6, 9, 10, 12, 13, 15, 16, 19, 42
already sorted, n = 9
median = 13
the 5th value
Q1 = 9.5, Q3 = 17.5
middles of each half
IQR = 8, 1.5 × 8 = 12
fence distance
fences: −2.5 and 29.5
9.5 − 12 and 17.5 + 12
A negative lower fence is perfectly normal. It simply means no value is low enough to be unusual.
This is why incomes and house prices are reported as medians. A handful of enormous values would pull a mean far above what most people actually experience.
The same fences; each case tests the edge of the rule.
A low outlier, from an even list: 2, 31, 34, 36, 38, 40, 41, 45
Q1 = 31 + 342 = 32.5, Q3 = 40 + 412 = 40.5
halves of four
IQR = 8, 1.5 × 8 = 12
the fence distance
fences: 20.5 and 52.5
2 is an outlier
Different: outliers can sit below the lower fence too — check both ends.
Exactly on the fence: Q1 = 14, Q3 = 26. Is 44 an outlier?
upper fence = 26 + 1.5 × 12 = 44
IQR = 12
44 is not beyond 44: not an outlier
only values past a fence count
Different: the fence itself is still inside; 44.1 would be an outlier, 44 is not.
Decimal data: newborn weights with Q1 = 2.8 kg and Q3 = 3.6 kg. Is 4.9 kg an outlier?
IQR = 0.8, 1.5 × 0.8 = 1.2
keep the units: kg
fences: 1.6 kg and 4.8 kg
2.8 − 1.2 and 3.6 + 1.2
4.9 > 4.8: an outlier
just past the upper fence
Different: small decimals make the margins thin — compare carefully, digit by digit.
Investigate before deleting anything.
| Possible cause | What to do |
|---|---|
| A recording error — 175 typed as 1750 | correct it |
| Someone outside the population | remove, and say so |
| A genuine extreme value | keep it — it is real data |
The 1.5 × IQR rule flags a value for attention. It does not give permission to delete it. Removing real data because it spoils a pattern is how honest statistics becomes dishonest.
Expect the IQR from Q1 and Q3, a lower or upper fence, "which value is an outlier" from a short list, and a question on which measure an outlier affects least. The fastest route is IQR, then 1.5 × IQR, then add it to Q3 and subtract it from Q1. The 10-second check: the fences must lie outside the box, and a value counts only if it is strictly beyond a fence.
Find the upper fence
A data set has Q1 = 24 and Q3 = 40. Above which value would a data point be called an outlier?
IQR = 40 − 24 = 16, and 1.5 × 16 = 24. The upper fence is Q3 + 24 = 64. D multiplies Q3 by 1.5 instead of the IQR. B adds only one IQR rather than one and a half.
Hint: work out the IQR as a number first, then multiply that by 1.5. Never multiply a quartile itself.
