Box Plots, Skewness and Outliers

Q: How do I know which way data is skewed?

Look at the longer tail. A longer right tail is positive skew; a longer left tail is negative skew.

Q: What is the 1,5 × IQR rule?

A value more than 1,5 IQRs below $Q_1$ or above $Q_3$ is treated as an outlier. You do not have to memorise it: the rule is given in the question when you need it.

Q: Should I use the mean or median for skewed data?

The median, because it is not pulled by extreme values.

The shape of a data set matters as much as its centre and spread. Box plots show the shape at a glance, and comparing the mean with the median tells you which way the data is skewed.

Five-number summary and box plots

The five-number summary is the minimum, $Q_{1}$ , median, $Q_{3}$ and maximum.

Example. $2; 4; 5; 5; 6; 7; 7; 8; 9; 10; 11; 12; 14; 15; 26$ (15 values)

Median: the 8th value, 8
$Q_{1}$ : median of the lower 7 values, 5
$Q_{3}$ : median of the upper 7 values, 12

Five-number summary: $(2; 5; 8; 12; 26)$ .

A box plot draws a box from $Q_{1}$ to $Q_{3}$ with a line at the median, and whiskers to the minimum and maximum.

Symmetric and skewed data

Symmetric: median in the middle of the box, equal whiskers. Mean $\approx$ median.
Positively skewed: long tail to the right. Median closer to $Q_{1}$ . Mean $>$ median.
Negatively skewed: long tail to the left. Median closer to $Q_{3}$ . Mean $<$ median.

Example. For the data above, the mean is $\frac{141}{15} = 9, 4$ , greater than the median of 8. The right whisker is much longer. The data is positively skewed.

Identifying outliers

An outlier lies far from the rest of the data. A standard test:

$x < Q_{1} - 1, 5 \times IQR or x > Q_{3} + 1, 5 \times IQR$

Example. IQR $= 12 - 5 = 7$ . Upper fence $= 12 + 10, 5 = 22, 5$ . The value 26 is an outlier.

Effect of an outlier. Removing 26 changes the mean from 9,4 to about 8,21, but the median only from 8 to 7,5. The mean is much more sensitive to outliers than the median.

Key terms

Five-number summary: Minimum, lower quartile, median, upper quartile and maximum.
Skewed: Not symmetric; with a longer tail on one side.
Outlier: A value much larger or smaller than the rest of the data.
Interquartile range: $Q_{3} - Q_{1}$ .

Common mistakes

Naming the skew after the side where data bunches up instead of the side of the long tail.
Including the median in both halves when finding quartiles for an odd number of values (follow your class convention consistently).
Using the range instead of the IQR in the outlier test.
Not ordering the data first.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

knowledge

Determine the five-number summary of:

$2; 4; 5; 5; 6; 7; 7; 8; 9; 10; 11; 12; 14; 15; 26$

Worked solution

Hidden - try the question first

Solution

Minimum 2, maximum 26
Median (8th value) $= 8$
$Q_{1} = 5$
$Q_{3} = 12$

Question 2

4 marks total

routine

For the same data:

(a) Calculate the IQR. (1)

(b) An outlier is a value greater than $Q_{3} + 1, 5 \times IQR$ or less than $Q_{1} - 1, 5 \times IQR$ . Show that 26 is an outlier. (3)

Worked solution

Hidden - try the question first

Solution

(a) $12 - 5 = 7$
(b) $Q_{3} + 1, 5 \times 7$
$= 22, 5$
$26 > 22, 5$ , so 26 is an outlier

Question 3

6 marks total

complex

For the same data:

(a) Calculate the mean. (2)

(b) Draw a box-and-whisker plot. (2)

Worked solution

Hidden - try the question first

Solution

(a) $12 - 5 = 7$
$= 9, 4$
(b) Box from 5 to 12 with the median at 8
Whiskers to 2 and 26 (or to 15 with 26 shown as an outlier)
(c) Positively skewed: mean $>$ median ( $9, 4 > 8$ )
and the median is closer to $Q_{1}$ than to $Q_{3}$ ( $3 < 4$ ), with a high outlier at 26

Question 4

8 marks total

problem-solving

The outlier 26 is removed from the data.

(a) Calculate the new mean. (2)

(b) Calculate the new median. (2)

(d) Which measure of centre would you use to describe the original data? Explain. (2)

Worked solution

Hidden - try the question first

Solution

(a) $12 - 5 = 7$
$\approx 8, 21$
(b) 14 values: median $= \frac{7 + 8}{2}$
$= 7, 5$
(c) The mean (from 9,4 to 8,21)
The mean uses every value, so one extreme value pulls it strongly
(d) The median
because it is not distorted by the outlier

FAQs

How do I know which way data is skewed?

Look at the longer tail. A longer right tail is positive skew; a longer left tail is negative skew.

What is the 1,5 × IQR rule?

A value more than 1,5 IQRs below $Q_{1}$ or above $Q_{3}$ is treated as an outlier. You do not have to memorise it: the rule is given in the question when you need it.

Should I use the mean or median for skewed data?

The median, because it is not pulled by extreme values.

Five-number summary and box plots

Symmetric and skewed data

Identifying outliers

Key terms

Common mistakes

Practice questions

FAQs

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