Open Math›Year 11›Mathematics Standard›Data analysis›Quartiles, box plots and outliers

Quartiles, box plots and outliers

Find quartiles and the interquartile range from ordered data, construct five-number summaries, identify outliers using fences, and interpret parallel box plots.

Worked examples

Finding quartiles and IQR

Straightforward

Problem

The number of hours of screen time recorded for 7 students, listed in order, are:

5, 9, 13, 17, 21, 25, 29

. Find

Q_1

Q_2

Q_3

and the IQR.

Count the values and locate the median (

Q_2

There are 7 values. The median is the middle (4th) value.

5, 9, 13, \mathbf{17}, 21, 25, 29

Q_2 = 17

Identify the lower half (values below

Q_2

) and find

Q_1

Lower half:

5, 9, 13

Q_1

is the middle value of the lower half:

Q_1 = 9

Identify the upper half (values above

Q_2

) and find

Q_3

Upper half:

21, 25, 29

Q_3

is the middle value of the upper half:

Q_3 = 25

Calculate the interquartile range.

\text{IQR} = Q_3 - Q_1 = 25 - 9 = 16

Answer

Q_1 = 9

Q_2 = 17

Q_3 = 25

, IQR

= 16

Five-number summary and outlier test

Moderate

Problem

A dataset of 8 values, listed in order, is:

10, 14, 18, 22, 26, 30, 34, 60

. Construct the five-number summary and determine whether 60 is an outlier.

Find the median (

Q_2

). With 8 values, it is the average of the 4th and 5th values.

Q_2 = \frac{22 + 26}{2} = 24

Find

Q_1

as the median of the lower half:

10, 14, 18, 22

Q_1 = \frac{14 + 18}{2} = 16

Find

Q_3

as the median of the upper half:

26, 30, 34, 60

Q_3 = \frac{30 + 34}{2} = 32

State the five-number summary.

\{\text{min},\ Q_1,\ Q_2,\ Q_3,\ \text{max}\} = \{10,\ 16,\ 24,\ 32,\ 60\}

Calculate the IQR and find the upper fence.

\text{IQR} = 32 - 16 = 16

\text{Upper fence} = Q_3 + 1.5 \times \text{IQR} = 32 + 1.5 \times 16 = 32 + 24 = 56

Compare the maximum value to the upper fence.

60 > 56

, so the value 60 lies beyond the upper fence. It is an outlier.

Answer

Five-number summary:

\{10, 16, 24, 32, 60\}

. The value 60 is an outlier (upper fence

= 56

Interpreting parallel box plots and the effect of removing an outlier

Challenging

Problem

Two datasets are displayed as parallel box plots with the following five-number summaries.

Dataset A: min

= 20

Q_1 = 35

Q_2 = 45

Q_3 = 65

, max

= 80

Dataset B: min

= 30

Q_1 = 50

Q_2 = 65

Q_3 = 75

, max

= 85

(a) Compare the centre, spread and shape of the two distributions.

(b) Dataset B consists of 6 values with a mean of 62. The value 85 is removed as an outlier. Find the mean of the remaining 5 values.

Compare the medians (centre).

Q_2

for Dataset A

= 45

;

Q_2

for Dataset B

= 65

.

Dataset B has a higher centre — a typical value in Dataset B is about 20 units higher than in Dataset A.

Compare the IQRs (spread of the middle 50%).

\text{IQR}_A = 65 - 35 = 30

\text{IQR}_B = 75 - 50 = 25

Dataset A has a greater IQR, so its middle 50% of values are more spread out.

Assess the shape of Dataset A.

Distance from

Q_1

Q_2

45 - 35 = 10

Distance from

Q_2

Q_3

65 - 45 = 20

The median is closer to

Q_1

, indicating a longer upper section. Dataset A is positively skewed.

Assess the shape of Dataset B.

Distance from

Q_1

Q_2

65 - 50 = 15

Distance from

Q_2

Q_3

75 - 65 = 10

The median is closer to

Q_3

, indicating a longer lower section. Dataset B is negatively skewed.

Find the total sum for Dataset B before removing the outlier.

\text{Sum} = n \times \bar{x} = 6 \times 62 = 372

Remove the outlier and find the new mean.

\text{New sum} = 372 - 85 = 287

\text{New mean} = \frac{287}{5} = 57.4

Removing the outlier of 85 reduces the mean from 62 to 57.4.

Answer

Dataset B has a higher median; Dataset A has a greater IQR. Dataset A is positively skewed; Dataset B is negatively skewed. After removing 85, the mean of Dataset B is

57.4

Practise

Q1·Straightforward

The data values below are listed in order.

3, 7, 9, 12, 15, 18, 21

Find the median (

Q_2

Explanation

There are 7 values. The median is the middle (4th) value:

Q_2 = 12

Q2·Straightforward

The data values below are listed in order.

5, 8, 11, 14, 17, 20, 23

Find the lower quartile (

Q_1

Explanation

There are 7 values. The median is the 4th value:

Q_2 = 14

. The lower half (values below the median) is

5, 8, 11

. The median of the lower half is the middle value:

Q_1 = 8

Q3·Straightforward

The data values below are listed in order.

2, 5, 8, 11, 14, 17

Find the interquartile range (IQR).

Explanation

There are 6 values. Split into lower half

2, 5, 8

and upper half

11, 14, 17

Q_1 = 5

(middle of lower half) and

Q_3 = 14

(middle of upper half).

\text{IQR} = 14 - 5 = 9

Q4·Straightforward

The data values below are listed in order.

4, 6, 8, 10, 12, 14, 16, 18

Find the upper quartile (

Q_3

Explanation

There are 8 values. Lower half:

4, 6, 8, 10

gives

Q_1 = \frac{6+8}{2} = 7

. Upper half:

12, 14, 16, 18

gives

Q_3 = \frac{14+16}{2} = 15

Q5·Moderate

The data values below are listed in order.

12, 15, 18, 21, 24, 27, 30

Find the interquartile range (IQR).

Explanation

There are 7 values.

Q_2 = 21

(4th value). Lower half:

12, 15, 18

gives

Q_1 = 15

. Upper half:

24, 27, 30

gives

Q_3 = 27

\text{IQR} = 27 - 15 = 12

Q6·Moderate

A dataset has lower quartile

Q_1 = 14

and upper quartile

Q_3 = 26

. Calculate the upper fence for identifying outliers using

Q_3 + 1.5 \times \text{IQR}

Explanation

\text{IQR} = Q_3 - Q_1 = 26 - 14 = 12

. Upper fence

= 26 + 1.5 \times 12 = 26 + 18 = 44

. Any value above 44 is an outlier.

Q7·Moderate

A dataset has

Q_1 = 8

and

Q_3 = 20

. A student claims that the value 42 is an outlier. Is the student correct?

Yes, because 42 is above the upper fence of 38No, because 42 is below the upper fence of 50Yes, because 42 is above the upper fence of 34No, because 42 equals the upper fence

Explanation

\text{IQR} = 20 - 8 = 12

. Upper fence

= 20 + 1.5 \times 12 = 20 + 18 = 38

. Since

42 > 38

, the value 42 is an outlier. The student is correct.

Q8·Moderate

A dataset has lower quartile

Q_1 = 35

and upper quartile

Q_3 = 55

. Calculate the lower fence for identifying outliers using

Q_1 - 1.5 \times \text{IQR}

Explanation

\text{IQR} = 55 - 35 = 20

. Lower fence

= 35 - 1.5 \times 20 = 35 - 30 = 5

. Any value below 5 is an outlier.

Q9·Challenging

Two classes sit a mathematics test. Their results are summarised below.

Class A:

Q_1 = 45

Q_2 = 60

Q_3 = 75

Class B:

Q_1 = 50

Q_2 = 68

Q_3 = 80

Which class has the higher median score?

Class AClass BBoth classes have the same medianCannot be determined from the information given

Explanation

Class A has

Q_2 = 60

and Class B has

Q_2 = 68

. Since

68 > 60

, Class B has the higher median score.

Q10·Challenging

Two datasets are summarised using five-number summaries.

Dataset P: min

= 10

Q_1 = 42

Q_2 = 55

Q_3 = 72

, max

= 90

Dataset Q: min

= 15

Q_1 = 50

Q_2 = 62

Q_3 = 74

, max

= 88

Which dataset has the greater interquartile range?

Dataset PDataset QBoth datasets have the same IQRCannot be determined

Explanation

\text{IQR}_P = 72 - 42 = 30

\text{IQR}_Q = 74 - 50 = 24

. Since

30 > 24

, Dataset P has the greater interquartile range, meaning the middle 50% of its values are more spread out.

Q11·Challenging

A box plot has

Q_1 = 20

Q_2 = 30

Q_3 = 55

. What does the position of the median suggest about the shape of the distribution?

Symmetric, because the median lies between Q1 and Q3Positively skewed, because the median is closer to Q1 than to Q3Negatively skewed, because the median is closer to Q1 than to Q3Positively skewed, because the median is closer to Q3 than to Q1

Explanation

Distance from

Q_1

Q_2

30 - 20 = 10

. Distance from

Q_2

Q_3

55 - 30 = 25

. The median is much closer to

Q_1

than to

Q_3

, indicating the upper part of the box is stretched. This suggests a longer upper tail — a positively skewed distribution.

Q12·Challenging

A dataset of 10 values has a mean of

12

. The value

48

is identified as an outlier and removed. Find the mean of the remaining 9 values.

Explanation

Total sum

= 10 \times 12 = 120

. After removing

48

: new sum

= 120 - 48 = 72

. New mean

= \frac{72}{9} = 8

. Removing the large outlier substantially lowers the mean.