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Outliers and the 1.5 IQR rule

One value sits far from all the others. How far is far enough to call it an outlier, and what should be done with it once it is found?

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When you reach for it

Heddwyn measures eleven young sharks that a research boat catches and releases. Ten are between 61 and 84 centimeters long; one is 131 centimeters. A value that sits far from the rest of the data is an outlier. An outlier can come about in three ways:

  • A mistake: a length misread, a digit typed wrong, a value written in the wrong unit.
  • A rare but real case: a shark that really is that long.
  • A member of a different group: an adult among young sharks, or an animal of another kind counted in by accident.

The numbers alone cannot say which happened; checking the record can. But first people must agree which values count as far, and by eye, on a dot plot, two people can disagree. The 1.5 IQR rule settles it with arithmetic, so everyone flags the same values. Reach for it whenever a data set seems to hold a value far from the rest, before reporting a summary of it or drawing a box plot that shows outliers.

The rule uses the Quartiles and the five number summary Q_1 and Q_3 and the interquartile range \text{IQR} = Q_3 - Q_1, found as in that lesson: sort the data, split it at the median, and when the count is odd leave the median out of both halves.

The procedure

The ruler is the IQR. The IQR is the width of the middle half of the data, the ordinary values, so measuring “far” in IQRs fits the rule to each data set: 10 millimeters beyond the middle half is far when that half spans 2 millimeters, and not far when it spans 50. The range would be a poor ruler, because the far value is itself one end of the range: the farther out it sits, the wider the range, so the ruler would grow with the very thing it measures. With six or more values the quartiles do not use the smallest or largest value, so the IQR stays put however far out an extreme value sits.

The fences are measured from the quartiles. Multiply the IQR by 1.5 to get the step. The lower fence is one step below Q_1 and the upper fence one step above Q_3:

\begin{aligned} \text{lower fence} &= Q_1 - 1.5 \times \text{IQR}\\ \text{upper fence} &= Q_3 + 1.5 \times \text{IQR} \end{aligned}

The steps start at the quartiles because the values from Q_1 to Q_3 are the middle half and are never in doubt; the question is how far a value has strayed beyond them. The 1.5 is a convention, agreed so that everyone gets the same answer; Box plots that show outliers shows what changes with 1 or 3 in its place.

Beyond a fence means an outlier. A value below the lower fence or above the upper fence is an outlier. A value exactly on a fence is not.

When the lower fence falls below zero. Lengths, weights and counts cannot be below zero. When the step is larger than Q_1, the lower fence lies below zero, no value can be below it, and there can be no low outliers in the data.

Then check every outlier. The rule flags a value; it does not judge it. What happens next depends on what the check finds:

  • a mistake whose right value is known is corrected, and the quartiles and fences are found again, because they came from a list with a wrong value in it;
  • a member of a different group is set aside when the question is about the group, and the report says so and why;
  • a rare but real value is kept, and the report says it is there.

No value is deleted just for being far. An outlier can also pull the Mean a long way while the median barely moves; Resistant and non resistant measures takes that up.

Worked examples

Example 1

One long shark

Heddwyn’s eleven sharks measure, in centimeters, 76, 68, 131, 72, 80, 61, 74, 84, 64, 78, 70. Use the 1.5 IQR rule to find any outliers.

Solution

Sort the lengths. There are 11, an odd count, so the median, the sixth value, is left out of both halves:

\begin{aligned} &\text{lower half: } 61,\ 64,\ 68,\ 70,\ 72\\ &\text{median: } 74\\ &\text{upper half: } 76,\ 78,\ 80,\ 84,\ 131 \end{aligned}

Each half has five values, so its quartile is its third value: Q_1 = 68 and Q_3 = 80. Then \text{IQR} = 12 and the step is 1.5 \times 12 = 18 centimeters. The fences are

68 - 18 = 50 \quad\text{and}\quad 80 + 18 = 98.

No shark is shorter than 50 centimeters, so there are no low outliers. The 131 centimeter shark is longer than 98, so it is an outlier; every other length lies between the fences (Figure 1). The 131 plays no part in Q_3, so it cannot drag the fences outward.

Heddwyn's 11 sharks lower fence upper fence 40 50 60 70 80 90 100 110 120 130 140 length in centimeters
Figure 1The fences sit one step of 18 centimeters beyond each quartile. Only the 131 centimeter shark lies outside them.
Example 2

A short beetle

Helga measures twelve beetles of one kind from a garden trap, in millimeters: 13.5, 15.9, 12.4, 17.2, 4.1, 14.6, 13.0, 16.4, 11.8, 15.3, 14.2, 15.0. Find any outliers.

Solution

Sort the lengths, six to a line:

\begin{aligned} &4.1,\ 11.8,\ 12.4,\ 13.0,\ 13.5,\ 14.2,\\ &14.6,\ 15.0,\ 15.3,\ 15.9,\ 16.4,\ 17.2. \end{aligned}

The count is even, so the two lines are the two halves, and each quartile is halfway between the third and fourth values of its half: \displaystyle Q_1 = \frac{12.4 + 13.0}{2} = 12.7 and \displaystyle Q_3 = \frac{15.3 + 15.9}{2} = 15.6. Then \text{IQR} = 2.9 millimeters and the step is 1.5 \times 2.9 = 4.35. The fences are 12.7 - 4.35 = 8.35 and 15.6 + 4.35 = 19.95.

The 4.1 millimeter beetle is below 8.35, so it is an outlier, while the longest, 17.2, is well inside the upper fence.

Example 3

When the lower fence falls below zero

Ignaz counts the fish he catches on ten trips: 5, 0, 7, 1, 16, 3, 1, 10, 2, 5. Are there any outliers?

Solution

Sorted: 0, 1, 1, 2, 3, 5, 5, 7, 10, 16. The halves are 0, 1, 1, 2, 3 and 5, 5, 7, 10, 16, so Q_1 = 1 and Q_3 = 7. Then \text{IQR} = 6 and the step is 1.5 \times 6 = 9.

The step of 9 is larger than Q_1 = 1, so the lower fence lies below zero. No trip catches fewer than 0 fish, so there can be no low outliers in this data.

The upper fence is 7 + 9 = 16. The trip with 16 fish sits exactly on the fence, and only a value above it is an outlier, so it is not one. This data set has no outliers.

Example 4

Two outliers, two stories

Hilario weighs the 13 melons in a crate of watermelons, in kilograms: 6.0, 21.0, 5.2, 7.5, 0.9, 6.4, 4.8, 7.2, 5.8, 6.1, 4.6, 6.8, 5.5. Find the outliers. On checking, the heaviest weighs the same again on a second scale, and the lightest is a cantaloupe packed into the crate by mistake. What should be done with each?

Solution

Thirteen is odd, so the seventh value, the median, is left out of both halves:

\begin{aligned} &\text{lower half: } 0.9,\ 4.6,\ 4.8,\ 5.2,\ 5.5,\ 5.8\\ &\text{median: } 6.0\\ &\text{upper half: } 6.1,\ 6.4,\ 6.8,\ 7.2,\ 7.5,\ 21.0 \end{aligned}

So \displaystyle Q_1 = \frac{4.8 + 5.2}{2} = 5.0 and \displaystyle Q_3 = \frac{6.8 + 7.2}{2} = 7.0. The IQR is 2 kilograms, the step 3, and the fences 5 - 3 = 2 and 7 + 3 = 10. The outliers are 0.9, below 2, and 21.0, above 10.

The 21.0 kilogram melon is rare but real: it belongs in the crate, so it stays, and a report on the crate says one melon was far heavier than the rest. The 0.9 kilogram fruit is from a different group, not a watermelon at all, so a summary of the watermelons leaves it out and says why. The check decides each case, not the distance: the melon that stays is the farther of the two.

Example 5

A mistake that hides an ordinary value

Ivanka writes down ten shark lengths in centimeters: 104, 1.1, 95, 120, 88, 107, 98, 115, 91, 102. Find any outliers. The tag sheet then shows the second shark was measured as 1.1 meters. Correct the data and look again.

Solution

Sorted: 1.1, 88, 91, 95, 98, 102, 104, 107, 115, 120. The halves have five values each, so Q_1 = 91 and Q_3 = 107. The IQR is 16, the step 24, and the fences are 91 - 24 = 67 and 107 + 24 = 131. The 1.1 is below 67, an outlier.

The tag sheet explains it: a length in meters written among lengths in centimeters. Since 1.1 meters is 110 centimeters, the right value is known and replaces the wrong one, and the fences are found again. Sorted, the lengths are now 88, 91, 95, 98, 102, 104, 107, 110, 115, 120, so Q_1 = 95 and Q_3 = 110. The IQR is 15, the step 22.5, and the fences are 95 - 22.5 = 72.5 and 110 + 22.5 = 132.5. Every length lies between them: no outliers. The flagged value was an ordinary shark, written down wrongly.

Example 6

How long would a new shark have to be?

Without the long shark, Heddwyn’s other ten measure 61, 64, 68, 70, 72, 74, 76, 78, 80, 84 centimeters. An eleventh shark is measured. Which lengths would make it an outlier among the eleven?

Solution

The fences come from all eleven values, so they depend on where the new length L lands in the sorted list.

Longer than all ten. If L is at least 84 it goes last. The median is 74, the halves are 61, 64, 68, 70, 72 and 76, 78, 80, 84, L, so Q_1 = 68 and Q_3 = 80 whatever L is. The step is 18 and the upper fence 98, so the shark is an outlier exactly when L is more than 98.

Shorter than all ten. If L is at most 61 it goes first. Now the median is 72, the halves are L, 61, 64, 68, 70 and 74, 76, 78, 80, 84, so Q_1 = 64 and Q_3 = 78. The IQR is 14, the step 21, and the lower fence 64 - 21 = 43: an outlier exactly when L is less than 43.

In between. For L from 61 to 84, Q_1 (the third value) is at most 68 and Q_3 (the ninth) at least 78, so the step is at least 1.5 \times 10 = 15. The fences lie at or beyond 68 - 15 = 53 and 78 + 15 = 93, so every such length is inside.

So a new shark is an outlier if it is shorter than 43 centimeters or longer than 98 centimeters.

Boundaries

Beyond means strictly beyond. A value exactly on a fence is not an outlier (Example 3).

The rule is one choice among several. Other multiples of the IQR and other quartile rules can flag different values, so a report says which rule it used.

Very small data sets. With few values a far value can help set a quartile and push its own fence away. In 1, 2, 3, 100 the halves are 1, 2 and 3, 100, so Q_1 = 1.5, Q_3 = 51.5 and the upper fence is 51.5 + 75 = 126.5: the 100 is not flagged, though it plainly sits far from the rest. The rule works as intended from six values up.

An outlier belongs to its data set. The fences come from the data, so a value far in one set can be ordinary in another, and adding, removing or correcting a value moves the fences (Examples 5 and 6).

Flagged is not wrong. The rule finds values to check, not values to delete. In a skewed data set several real values in the long tail can be flagged at once.

Where this goes next

A box plot draws the five-number summary; a modified box plot also draws the outliers found here as dots of their own. Describing a data set in words (Describing a distribution) names any outliers alongside its center, spread and shape. With two variables, a point can be far from the pattern in a new way: Outliers, leverage and influential points.

Exercises

Fences from the quartiles

  1. A data set has Q_1 = 20 and Q_3 = 28. Find its fences. Is 9 an outlier? Is 41?
  2. A data set has the five-number summary 14, 22, 26, 30, 47. Find its fences. Is the minimum an outlier? Is the maximum?
  3. Beetle lengths have Q_1 = 3.5 millimeters and Q_3 = 6.5 millimeters. Find the upper fence and say what happens at the lower end. Which of 0.5, 10.8, 11 and 12.4 millimeters would be outliers?
  4. The numbers of fish caught on some trips have Q_1 = 2 and Q_3 = 10. Which whole numbers of fish would be outliers?

Outliers in a list

For each data set find Q_1, Q_3, the fences and any outliers.

  1. Ivanka’s nine sharks, in centimeters: 104, 97, 112, 99, 61, 108, 101, 95, 110.
  2. Helga’s ten beetles, in millimeters: 8.2, 7.5, 9.0, 8.8, 7.9, 12.6, 8.4, 8.1, 7.7, 8.6.
  3. Ignaz’s catches on twelve trips: 3, 0, 8, 2, 23, 5, 1, 9, 0, 6, 2, 4.
  4. Hilario’s eleven watermelons, in kilograms: 7.4, 2.1, 8.3, 6.9, 13.5, 7.8, 6.4, 8.7, 7.1, 8.0, 7.6.
  5. Thirteen ferry crossings, in minutes: 31, 24, 48, 28, 36, 20, 33, 41, 25, 38, 27, 35, 30.

From a table or a dot plot

Table A shows how many beetles Helga found under each of 20 logs.

Beetles under a log01234514
Logs4543211
  1. From Table A, find the quartiles, the fences and any outliers.
  2. Figure 2 shows the lengths of 15 beetles. Find the fences and any outliers.
15 beetles 0 2 4 6 8 10 12 14 16 18 20 length in millimeters
Figure 2The lengths of 15 beetles, for Exercise 11.

Causes and choices

  1. Each value below was flagged by the rule. Say which cause is most likely and what should be done. (a) A survey of one kind of beetle lists one of 41 millimeters among lengths from 8 to 13; the notes say it was a stag beetle. (b) A ferry crossing took 75 minutes; the log says a storm held it in the harbor for 40 of them. (c) A shark’s length is written as 9.8 among lengths from 85 to 120 centimeters; a photo of the shark on the measuring board shows 98.
  2. In Exercise 5, the 61 turns out to be a typing slip for 106. Find the fences again. Are there any outliers now?
  3. A data set has an IQR of 12 and an upper fence of 70. Find Q_3, Q_1 and the lower fence.
  4. Ten ferry crossings took 30, 32, 35, 36, 38, 40, 41, 43, 45, 47 minutes. An eleventh takes longer than all ten. What is the shortest whole number of minutes that would make it an outlier among the eleven?

Explain your reasoning

  1. Explain why the rule measures distance in IQRs rather than in ranges.
  2. Can more than half of the values in a data set be outliers? Explain.
  3. A report on a shark survey leaves out the longest shark without saying so. Explain how a reader could be misled, and what the report should do instead.
  4. A data set of counts has Q_1 = 4 and Q_3 = 8. Can a count of 0 be an outlier? Explain.
Answers
  1. Fences 8 and 40. The IQR is 8 and the step 12. The 9 is above 8, so it is not an outlier; 41 is above 40, so it is.
  2. Fences 10 and 42. The IQR is 30 - 22 = 8 and the step 12. The minimum 14 is not an outlier; the maximum 47 is.
  3. Upper fence 11 millimeters, no low outliers, and only 12.4 is an outlier. The step is 4.5, larger than Q_1 = 3.5, so the lower fence lies below zero. The upper fence is 6.5 + 4.5 = 11, and 11 itself is on the fence, so it is not an outlier.
  4. 23 fish or more. The step is 12, larger than Q_1 = 2, so there are no low outliers. The upper fence is 10 + 12 = 22.
  5. Q_1 = 96, Q_3 = 109, fences 76.5 and 128.5; the outlier is 61. Sorted: 61, 95, 97, 99, 101, 104, 108, 110, 112; the median 101 is left out, so \displaystyle Q_1 = \frac{95 + 97}{2} and \displaystyle Q_3 = \frac{108 + 110}{2}. The IQR is 13, the step 19.5.
  6. Q_1 = 7.9, Q_3 = 8.8, fences 6.55 and 10.15; the outlier is 12.6. Sorted: 7.5, 7.7, 7.9, 8.1, 8.2, 8.4, 8.6, 8.8, 9.0, 12.6. The IQR is 0.9, the step 1.35.
  7. Q_1 = 1.5, Q_3 = 7, upper fence 15.25, no low outliers; the outlier is 23. Sorted: 0, 0, 1, 2, 2, 3, 4, 5, 6, 8, 9, 23, so \displaystyle Q_1 = \frac{1 + 2}{2} and \displaystyle Q_3 = \frac{6 + 8}{2}. The step is 8.25, larger than Q_1.
  8. Q_1 = 6.9, Q_3 = 8.3, fences 4.8 and 10.4; the outliers are 2.1 and 13.5. Sorted: 2.1, 6.4, 6.9, 7.1, 7.4, 7.6, 7.8, 8.0, 8.3, 8.7, 13.5; the median 7.6 is left out. The IQR is 1.4, the step 2.1.
  9. Q_1 = 26, Q_3 = 37, fences 9.5 and 53.5; no outliers. Sorted: 20, 24, 25, 27, 28, 30, 31, 33, 35, 36, 38, 41, 48; the median 31 is left out, so \displaystyle Q_1 = \frac{25 + 27}{2} and \displaystyle Q_3 = \frac{36 + 38}{2}. The step is 16.5.
  10. Q_1 = 1, Q_3 = 3, upper fence 6, no low outliers; the outlier is 14. The running totals are 4, 9, 13, 16, 18, 19, 20. The lower half is positions 1 to 10, whose 5th and 6th values are both 1; the upper half’s 15th and 16th values are both 3. The step is 3, larger than Q_1.
  11. Fences 7 and 15; the outliers are 3 and 18. The running totals are 1, 3, 6, 10, 13, 14, 15. The median, the 8th value, is left out; the halves’ middles are the 4th value, 10, and the 12th, 12. The step is 3.
  12. (a) A different group: a stag beetle is not the kind surveyed, so it is set aside and the report says so. (b) Rare but real: the crossing really took that long, so it is kept and the storm is mentioned. (c) A mistake: the right value is known, so it becomes 98 centimeters and the quartiles and fences are found again.
  13. Fences 81.5 and 125.5; no outliers. Sorted: 95, 97, 99, 101, 104, 106, 108, 110, 112, so \displaystyle Q_1 = \frac{97 + 99}{2} = 98 and \displaystyle Q_3 = \frac{108 + 110}{2} = 109. The step is 16.5.
  14. Q_3 = 52, Q_1 = 40, lower fence 22. The step is 1.5 \times 12 = 18, so Q_3 = 70 - 18, Q_1 = 52 - 12, and the lower fence is 40 - 18.
  15. 61 minutes. With the new time last, the median 40 is left out and the halves are 30, 32, 35, 36, 38 and 41, 43, 45, 47 with the new time, so Q_1 = 35 and Q_3 = 45 whatever it is. The step is 15 and the upper fence 60.
  16. The far value is one end of the range, so the farther out it sits the wider the range, and a fence measured in ranges would move out with it. The IQR comes from the middle half, which the far value does not touch, so it measures the spread of the ordinary values.
  17. No, at most half. Every outlier lies below Q_1 or above Q_3. The values below Q_1 are at most half of the lower half, and those above Q_3 at most half of the upper half, so together they are at most half of the data.
  18. A reader would take the summary to describe every shark caught, never learning that one was far longer. If the shark was real, the report keeps it and says so; if it was set aside, for a mistake or another group, the report says that and why.
  19. No. The IQR is 4 and the step 6, larger than Q_1 = 4, so the lower fence lies below zero. A count of 0 is not below it, and there can be no low outliers in this data.
kg1128 · updated 2026-10-13 · CC-BY-SA-4.0