Measures of Dispersion

Lessons

Senior 4 · Mathematics

Measures of Dispersion

3 lessons

Lesson 1 of 3

Central tendency, quartiles and box plots

25 min

Same average, different journeys

Two bus companies run the Kigali to Huye route from Nyabugogo. A passenger timed 11 trips with each company. The times, in minutes and in order, were (illustrative data):

Company	Journey times (minutes)
A	142, 145, 147, 148, 149, 150, 151, 152, 153, 155, 158
B	120, 128, 135, 140, 146, 150, 156, 160, 165, 172, 178

Both lists add up to $1650$ , so both companies have a mean of $\frac{1650}{11} = 150$ minutes, and both have a median of 150 minutes. An average alone says the two services are the same. They are not: with Company B you might arrive half an hour early or half an hour late. If you have a 13:00 appointment in Huye, the spread of the times matters as much as their centre. This unit is about measuring that spread. This first lesson revises the measures of centre and introduces the quartiles and the box plot, the picture that shows centre and spread together.

Definition

Measures of central tendency: single values that describe the centre of a data set $x_1, x_2, \dots, x_n$ .

The mean $\bar{x}$ (read "x bar") is the sum of the values divided by their number: $\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i$ . For a frequency table with values $x_{i}$ and frequencies $f_{i}$ , $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ .
The median is the middle value once the data are in order: the $\frac{n+1}{2}$ th value.
The mode is the value with the greatest frequency. A data set may have one mode, several, or none.

Example: for 2, 3, 3, 7, 10: $\bar{x} = \frac{25}{5} = 5$ , median = 3rd value = 3, mode = 3.

Worked example 1: centre of a frequency table

A survey asked 40 households in one village how many children aged under 5 live there (illustrative data). Find the mean, the median and the mode.

Children, $x$	0	1	2	3	4	5
Households, $f$	3	7	12	10	5	3
$f x$	0	7	24	30	20	15
Cumulative $f$	3	10	22	32	37	40

Solution

Mean: $\bar{x} = \frac{\sum fx}{\sum f} = \frac{96}{40} = 2.4$ children.

Median: $n = 40$ , so the median lies halfway between the 20th and 21st values. The cumulative frequencies show that the 11th to the 22nd values are all 2, so the median is 2.

Mode: the greatest frequency is 12, for $x = 2$ , so the mode is 2.

The mean 2.4 is not a possible number of children. That is normal: a mean describes the whole group, not one household.

Quartiles: cutting the data into four

The median splits ordered data into two halves. The quartiles split it into four quarters, and the distance between the outer two quartiles measures how spread out the middle half of the data is.

Definition

Quartiles: for $n$ values in increasing order,

Q_1 = \text{the } \tfrac{n+1}{4}\text{th value}, \qquad Q_2 = \text{the } \tfrac{n+1}{2}\text{th value (the median)}, \qquad Q_3 = \text{the } \tfrac{3(n+1)}{4}\text{th value}.

$Q_{1}$ is the lower quartile and $Q_{3}$ the upper quartile.

Measures of dispersion describe how far the data spread out. The simplest two are the range $= \text{maximum} - \text{minimum}$ and the interquartile range $\text{IQR} = Q_3 - Q_1$ .

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

Example: Company A above has $n = 11$ , so $Q_{1}$ is the 3rd value, 147, and $Q_{3}$ is the 9th value, 153. Its IQR is $153 - 147 = 6$ minutes.

Worked example 2: a position that is not a whole number

The numbers of customers served in one hour at a mobile money kiosk on 10 days were 3, 5, 6, 8, 9, 11, 12, 14, 15, 20. Find the quartiles and the IQR.

Solution

$n = 10$ , so $n + 1 = 11$ .

$Q_{1}$ : position $\frac{11}{4} = 2.75$ . Go three quarters of the way from the 2nd value (5) to the 3rd (6): $Q_{1} = 5 + 0.75 (6 - 5) = 5.75$ .
Median: position $5.5$ , halfway between 9 and 11, so $Q_{2} = 10$ .
$Q_{3}$ : position $\frac{33}{4} = 8.25$ . Go a quarter of the way from the 8th value (14) to the 9th (15): $Q_{3} = 14.25$ .

$\text{IQR} = 14.25 - 5.75 = 8.5$ customers.

Warning

Books and calculators do not all find quartiles the same way. Some take the medians of the lower and upper halves; some spreadsheets use yet another position rule. For large data sets the answers are almost equal, but for 10 values they can differ by a little. In an exam, use one method, show the positions you used, and you will be marked right. In this unit we always use the $\frac{n+1}{4}$ rule above.

The box plot

A box plot (or box-and-whisker diagram) draws the five-number summary on a scale. The box runs from $Q_{1}$ to $Q_{3}$ , with a line at the median; the whiskers run out to the smallest and largest values. Here are the two bus companies on one scale:

The picture says at once what the table hid. The medians are the same, but Company B's box is five times as wide (IQR 30 minutes against 6), and its whiskers stretch over 58 minutes against 16. Company A is far more consistent.

Outliers and the shape of the data

An outlier is a value that lies unusually far from the rest. A common rule marks as an outlier any value below $Q_1 - 1.5 \times \text{IQR}$ or above $Q_3 + 1.5 \times \text{IQR}$ . These two numbers are called the fences. On a box plot an outlier is drawn as a separate cross, and the whisker stops at the most extreme value inside the fences.

The box plot also shows skewness. If the median sits near $Q_{1}$ and the right whisker is long, the data are positively skewed (a tail of large values, as with incomes). If the median sits near $Q_{3}$ and the left whisker is long, they are negatively skewed. If the two halves balance, the data are roughly symmetric.

Worked example 3: finding an outlier

The marks out of 50 of 15 students in a test were: 12, 18, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 35, 48. Draw a box plot and decide whether any mark is an outlier.

Solution

$n = 15$ , so $n + 1 = 16$ : $Q_{1}$ is the 4th value, 23; the median is the 8th value, 27; $Q_{3}$ is the 12th value, 31. So $\text{IQR} = 31 - 23 = 8$ and $1.5 \times \text{IQR} = 12$ .

\text{lower fence} = 23 - 12 = 11, \qquad \text{upper fence} = 31 + 12 = 43.

48 is above 43, so 48 is an outlier. 12 is above 11, so it is not (only just). The right whisker stops at 35, the largest mark inside the fences:

Before deleting an outlier, ask why it is there. A mark of 48 is not a mistake: it is one strong student. An outlier caused by a typing error (480 for 48) should be corrected; a real one should be kept and reported.

Tip

To compare two groups, draw their box plots on the same scale, one above the other, and then write two sentences: one comparing the medians (which is typically higher) and one comparing the IQRs (which is more consistent). That is exactly what an examiner wants to read.

Mario's Math Tutoring: Outliers, Box and Whisker Plot (1.5 IQR). Watch how the fences are found from the quartiles and how an outlier is drawn apart from the whisker.

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Practice

Try each one before opening the answers. They get harder as you go.

Find the mean, median and mode of 4, 9, 6, 4, 12, 7, 4, 10.
In a survey, 25 students gave the number of meals they ate the day before: 1 meal (2 students), 2 meals (9), 3 meals (12), 4 meals (2). Find the mean and the median.
Find $Q_{1}$ , $Q_{3}$ and the IQR of: 31, 35, 36, 40, 42, 44, 47, 50, 52, 55, 61.
Find the quartiles of 2, 4, 4, 5, 7, 8, 9, 13 using the $\frac{n+1}{4}$ rule.
A data set has $Q_{1} = 40$ and $Q_{3} = 52$ . Find the fences. Are 20 and 70 outliers?
The heights, in cm, of 11 seedlings were: 5, 9, 10, 11, 12, 12, 13, 14, 15, 16, 30. Give the five-number summary, decide which values are outliers, and say where the right whisker of the box plot should end.
Two box plots have the same median. Plot P has IQR 4 and range 30; plot Q has IQR 10 and range 14. Which set has the more spread-out middle half? What does P's large range suggest?

Show answersHide answers

Ordered: 4, 4, 4, 6, 7, 9, 10, 12. Mean $\frac{56}{8} = 7$ ; median $\frac{6 + 7}{2} = 6.5$ ; mode 4.
$\sum fx = 2 + 18 + 36 + 8 = 64$ , mean $\frac{64}{25} = 2.56$ meals. Median = 13th value; cumulative frequencies 2, 11, 23, so the median is 3 meals.
$n = 11$ : $Q_{1}$ = 3rd value = 36, $Q_{3}$ = 9th value = 52, IQR = 16.
$n = 8$ : $Q_{1}$ at position 2.25: $4 + 0.25 (4 - 4) = 4$ . Median at 4.5: $6$ . $Q_{3}$ at 6.75: $8 + 0.75 (9 - 8) = 8.75$ .
IQR = 12, $1.5 \times 12 = 18$ . Fences: $40 - 18 = 22$ and $52 + 18 = 70$ . 20 is below 22, so it is an outlier; 70 is not above 70, so it is not.
Minimum 5, $Q_{1} = 10$ , median 12, $Q_{3} = 15$ , maximum 30. IQR = 5, fences $2.5$ and $22.5$ : 30 is an outlier (5 is not). The right whisker ends at 16, the largest value inside the fences, and 30 is drawn as a separate cross.
Q has the more spread-out middle half (IQR 10 against 4). P's range of 30 is large compared with its IQR, which suggests one or more extreme values (possible outliers) far from the centre.

Quick check

Twelve values are in increasing order. Using the (n + 1)/4 rule, where is the lower quartile Q₁?

Key points

Mean $\bar{x} = \frac{\sum f x}{\sum f}$ ; median = $\frac{n+1}{2}$ th value; mode = most frequent value.
$Q_{1}$ and $Q_{3}$ are the $\frac{n+1}{4}$ th and $\frac{3(n+1)}{4}$ th values; $\text{IQR} = Q_3 - Q_1$ .
A box plot draws the five-number summary; values beyond $Q_1 - 1.5\,\text{IQR}$ or $Q_3 + 1.5\,\text{IQR}$ are outliers.
Compare groups by their medians (centre) and their IQRs (spread).

Sources and credits

Written by Ganzaa: the explanations, the diagrams, the worked examples, the practice items and the questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced. The journey times, household and test data are illustrative.
Photo: Nyabugogo Bus Station, Francisco Anzola, Wikimedia Commons, CC BY 3.0.
Video: Outliers - Box and Whisker Plot (1.5 IQR), Mario's Math Tutoring, embedded from YouTube.

1Same average, different journeys

2Quartiles: cutting the data into four

3The box plot

4Outliers and the shape of the data

Same average, different journeys

Quartiles: cutting the data into four

The box plot

Outliers and the shape of the data