Outliers and the 1.5 × IQR Rule

The 1.5 × IQR rule is the standard way to find outliers. Find the interquartile range, IQR = Q3 − Q1, and multiply it by 1.5. Any value below Q1 − 1.5 × IQR or above Q3 + 1.5 × IQR is an outlier. These two cutoffs are called the lower and upper fences.

Before this lesson: Box Plots, Describing a Distribution and Standard Deviation

What makes a value an outlier

An outlier is a value that falls far outside the overall pattern of the data. Looking at a graph is a good start, but people can disagree about what “far” means. The 1.5 × IQR rule gives everyone the same test.

The rule measures distance in IQRs. The IQR is the width of the middle half of the data, so a value more than one and a half IQRs past the box is unusually far out. Because the quartiles are resistant, the outliers themselves don’t change the fences much.

The 1.5 × IQR rule

\text{IQR} = Q_3 - Q_1

\text{Lower fence} = Q_1 - 1.5 \times \text{IQR} \qquad \text{Upper fence} = Q_3 + 1.5 \times \text{IQR}

1. Put the data in order and find the median.
2. Find Q1, the median of the lower half, and Q3, the median of the upper half. (When n is odd, leave the median out of both halves.)
3. Find the IQR = Q3 − Q1, then 1.5 × IQR.
4. Find the fences: subtract 1.5 × IQR from Q1 and add it to Q3.
5. Any value below the lower fence or above the upper fence is an outlier.

Outlier examples

Example 1: An outlier on the high side

Find any outliers in these 11 values (example data): 12, 15, 17, 18, 19, 20, 21, 22, 24, 25, 41.

\text{Median} = 20

Q_1 = 17

Q_3 = 24

\text{IQR} = 24 - 17 = 7

1.5 \times 7 = 10.5

17 - 10.5 = 6.5

24 + 10.5 = 34.5

41 \gt 34.5

Answer: 41 is an outlier. No value is below the lower fence.

Example 2: An outlier on the low side

Find any outliers in these 10 values (example data): 3, 28, 31, 33, 34, 36, 37, 39, 40, 42.

\text{Median} = \dfrac{34 + 36}{2} = 35

Q_1 = 31

Q_3 = 39

\text{IQR} = 39 - 31 = 8

1.5 \times 8 = 12

31 - 12 = 19, \; 39 + 12 = 51

3 \lt 19

Answer: 3 is an outlier.

Example 3: A value that looks far but isn’t an outlier

Do these 8 values have an outlier (example data)? 5, 8, 9, 10, 12, 14, 15, 22.

Q_1 = \dfrac{8 + 9}{2} = 8.5

Q_3 = \dfrac{14 + 15}{2} = 14.5

\text{IQR} = 14.5 - 8.5 = 6

1.5 \times 6 = 9

8.5 - 9 = -0.5, \; 14.5 + 9 = 23.5

22 \lt 23.5

Answer: no outliers. 22 is 7 above the next value, but it is still within the fences, so the rule does not flag it.

Example 4: Screen time and a modified box plot (word problem)

Fifteen teens record their daily screen time in minutes (example data): 95, 110, 120, 125, 130, 135, 140, 145, 150, 155, 160, 170, 180, 190, 320. Find any outliers and describe the modified box plot.

\text{Median} = 145

Q_1 = 125

Q_3 = 170

\text{IQR} = 170 - 125 = 45

1.5 \times 45 = 67.5

125 - 67.5 = 57.5, \; 170 + 67.5 = 237.5

320 \gt 237.5

On a TI-84: enter the data in L1, then STAT, CALC, 1:1-Var Stats to see Q1 = 125, Med = 145 and Q3 = 170. For the graph, press 2nd, STAT PLOT, turn Plot1 on, and choose the first box-plot icon (the one with separate dots). ZOOM, 9:ZoomStat draws it; TRACE shows the outlier at 320. In Desmos: type s = [95, 110, 120, 125, 130, 135, 140, 145, 150, 155, 160, 170, 180, 190, 320], then boxplot(s). Some software finds quartiles a slightly different way, so its Q1 and Q3 can differ a little from hand values.

Answer: 320 minutes is an outlier. In the modified box plot (shown above), the box runs from 125 to 170, the whiskers reach 95 and 190 (the most extreme values inside the fences), and 320 is a separate dot.

Example 5: How an outlier affects the summary

Use the data from Example 2. Compare the mean and median with and without the outlier, 3.

3 + 28 + 31 + 33 + 34 + 36 + 37 + 39 + 40 + 42 = 323

\bar{x} = \dfrac{323}{10} = 32.3

323 - 3 = 320

\bar{x} = \dfrac{320}{9} \approx 35.6

\text{Median without 3} = 36

Answer: removing 3 moves the mean by about 3.3 (32.3 to 35.6) but the median by only 1 (35 to 36). The median is resistant. Never delete an outlier just because it is unusual: check whether it is a mistake, and report the median and IQR if it is real.

Common mistakes

Related methods

Other outlier rules

For roughly symmetric data, some people flag values more than 2 standard deviations from the mean, using the standard deviation. The 1.5 × IQR rule works for any shape, because it does not use the mean. Values beyond 3 × IQR from the box are sometimes called extreme outliers.

Outliers in a full description

An outlier is the “O” in SOCS, so mention it whenever you describe a distribution. Percentiles and z-scores, the next lesson, give another way to say how unusual a value is.

Practice problems

1. A data set has Q1 = 20 and Q3 = 30. Find the lower and upper fences.
2. A data set has Q1 = 40 and Q3 = 52. Is 75 an outlier?
3. Find any outliers: 2, 4, 5, 6, 7, 8, 9, 30.
4. Find any outliers: 1, 20, 22, 23, 25, 26, 27.
5. Find any outliers: 10, 12, 13, 15, 16, 18, 20.
6. A five-number summary is min = 4, Q1 = 10, median = 14, Q3 = 18, max = 35. Is the minimum an outlier? Is the maximum?
7. A data set has mean 50 and standard deviation 6. Is 64 more than 2 standard deviations from the mean?
8. A survey of hourly pay at a store includes one value of $250 per hour. What should you do before removing it?
9. Find any outliers: 15, 48, 50, 51, 53, 54, 55, 57, 58, 90.
10. For 4, 5, 6, 7, 28, find the mean and median with and without 28.

\text{IQR} = 30 - 20 = 10

1.5 \times 10 = 15

20 - 15 = 5, \; 30 + 15 = 45

\text{IQR} = 52 - 40 = 12

1.5 \times 12 = 18

52 + 18 = 70

75 \gt 70

Q_1 = \dfrac{4 + 5}{2} = 4.5, \; Q_3 = \dfrac{8 + 9}{2} = 8.5

\text{IQR} = 8.5 - 4.5 = 4

1.5 \times 4 = 6

4.5 - 6 = -1.5, \; 8.5 + 6 = 14.5

30 \gt 14.5

Q_1 = 20, \; Q_3 = 26

\text{IQR} = 26 - 20 = 6

1.5 \times 6 = 9

20 - 9 = 11, \; 26 + 9 = 35

1 \lt 11

Q_1 = 12, \; Q_3 = 18

\text{IQR} = 18 - 12 = 6

12 - 9 = 3, \; 18 + 9 = 27

3 \lt 10, \; 20 \lt 27

\text{IQR} = 18 - 10 = 8

1.5 \times 8 = 12

10 - 12 = -2, \; 18 + 12 = 30

4 \gt -2, \; 35 \gt 30

50 + 2(6) = 62

64 \gt 62

\text{Check the source}

\text{Median} = \dfrac{53 + 54}{2} = 53.5

Q_1 = 50, \; Q_3 = 57

\text{IQR} = 57 - 50 = 7

1.5 \times 7 = 10.5

50 - 10.5 = 39.5, \; 57 + 10.5 = 67.5

15 \lt 39.5, \; 90 \gt 67.5

\bar{x} = \dfrac{4 + 5 + 6 + 7 + 28}{5} = \dfrac{50}{5} = 10

\text{Median} = 6

\bar{x} = \dfrac{4 + 5 + 6 + 7}{4} = \dfrac{22}{4} = 5.5

\text{Median} = \dfrac{5 + 6}{2} = 5.5

Outliers and the 1.5 × IQR rule FAQ

What is the 1.5 IQR rule?

It says a value is an outlier if it is more than 1.5 × IQR below Q1 or more than 1.5 × IQR above Q3. The cutoffs Q1 − 1.5 × IQR and Q3 + 1.5 × IQR are called fences.

How do you find outliers in a data set?

Find Q1, Q3 and the IQR, then compute the fences. Check the smallest and largest values: any value outside the fences is an outlier.

What is a modified box plot?

It is a box plot that shows outliers as separate dots. Its whiskers stop at the most extreme values that are still inside the fences.

Should you remove outliers?

Not automatically. Remove or correct an outlier only if it is a mistake, such as a typing error. A real outlier is part of the data, so keep it and use resistant measures like the median and IQR.

Summary

• IQR = Q3 − Q1.
• Fences: Q1 − 1.5 × IQR and Q3 + 1.5 × IQR.
• A value below the lower fence or above the upper fence is an outlier.
• A modified box plot shows outliers as dots, with whiskers to the last values inside the fences.
• Investigate outliers before removing them; the median and IQR resist them, the mean and standard deviation do not.

Previous lesson: Standard Deviation

Next lesson: Percentiles and Z-Scores

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