Measures of Central Tendency

Look at your Grade 8 gradesheet, or at this one: the grade points in Mathematics of nine friends of Grade 8.

2.8, 3.6, 2.4, 3.2, 4.0, 2.8, 3.6, 2.8, 3.2

Three questions, three different answers: 3.16, 3.2 and 2.8. Each is a fair way of saying “a typical grade point”, and each answers a slightly different question. Numbers like these, which say where the middle of the data is, are called measures of central tendency. You met data in tables in Unit 17; here we find their middle.

Activity 1: 32 marks

a. The marks are written one by one, a value for each student: an individual series. Add them all. $\sum$ (the Greek capital letter sigma) means “the sum of”, and X stands for each mark:

$$\sum X = 9 + 9 + 10 + \cdots + 18 = 448$$

There are N = 32 marks. Sharing the 448 marks equally among the 32 students gives each

$$\overline{X} = \frac{\sum X}{N} = \frac{448}{32} = 14$$

X (read “X bar”) is the arithmetic mean, also called the mean or the average: the total shared out equally.

b. Adding 32 numbers one by one is slow, and many of them repeat. Count how many students got each mark, its frequency f:

Marks	9	10	12	14	16	18
Students	2	2	7	9	8	4

A table of values with their frequencies is a discrete series.

c. Seven students got 12: together that is 7 × 12 = 84 marks. So multiply each mark by its frequency, fX, and add:

Marks (X)	Students (f)	fX
9	2	18
10	2	20
12	7	84
14	9	126
16	8	128
18	4	72
	$N = \sum f = 32$	$\sum fX = 448$

$$\overline{X} = \frac{\sum fX}{\sum f} = \frac{\sum fX}{N} = \frac{448}{32} = 14$$

The same sum, 448, and the same mean, 14, with six multiplications instead of 31 additions.

The arithmetic mean of data is the sum of all the values divided by the number of values. For an individual series $\overline{X} = \frac{\sum X}{N}$; for a discrete series $\overline{X} = \frac{\sum fX}{N}$, where $N = \sum f$ is the total of the frequencies. When there are many values and they repeat, write them as a discrete series first: it is quicker.

Worked examples: the mean

Activity 2: the student in the middle

At home, line up your family by height, or by age, and find the median the same way.

Activity 3: ages of prize winners

a. Each age is written separately: an individual series.

b. In ascending order (smallest first):

8, 8, 8, 10, 10, 10, 10, 10, 12, 12, 12, 12, 12, 12, 12, 12, 12, 14, 14, 14, 14, 14, 14, 14, 14, 16, 16, 16, 18

c. There are 29 ages. The 15th has 14 ages before it and 14 after it, so it is exactly in the middle. Count to it: the 15th age is 12. There is a rule for the place of the middle item:

$$\begin{aligned} \text{Median} &= \left(\frac{N + 1}{2}\right)\text{th item} \\ &= \left(\frac{29 + 1}{2}\right)\text{th item} = 15\text{th item} \end{aligned}$$

d. The middle value is the median, written $M_d$. Here $M_d = 12$ years.

e. As a discrete series:

Age (years)	8	10	12	14	16	18
Students	3	5	9	8	3	1

The cumulative frequency (cf) of a value is how many items there are up to and including it: add the frequencies one by one.

Age (years)	Frequency (f)	Cumulative frequency (cf)
8	3	3
10	5	3 + 5 = 8
12	9	8 + 9 = 17
14	8	17 + 8 = 25
16	3	25 + 3 = 28
18	1	28 + 1 = 29
	N = 29

f. The median is the 15th item, as in c. Read the cf column: items 1 to 3 are 8, items 4 to 8 are 10, items 9 to 17 are 12. The 15th is among them, so $M_d = 12$, without writing all 29 ages out.

The median is the value that divides data, arranged in order, into two equal parts: half the values are at or below it and half at or above it. It is written $M_d$. Its place is

$$\text{Median} = \left(\frac{N + 1}{2}\right)\text{th item}$$

We divide by 2 because the median cuts the data in two. In a discrete series, put the values in ascending (or descending) order, make the cumulative frequencies, and find the first cf that reaches the median’s place: its value is the median.

Worked examples: the median

Activity 4: hours of study

a. As a discrete series:

Time (hours)	8	10	12	13	15	17	18
Frequency (f)	4	3	2	5	3	3	1

b. 13 hours comes 5 times, more than any other value.

c. So the mode is 13 hours: $M_o = 13$.

The mode is the value that occurs most often, the one with the highest frequency. It is written $M_o$:

$$M_o = \text{the value with the highest frequency}$$

No calculation is needed, only counting. A set of data can have two modes (it is then bimodal), or none at all when every value occurs equally often, like Kathmandu’s 12 monthly rainfalls, which are all different.

Which one to use?

The three can tell different stories about the same data, so the choice matters.

	Mean	Median	Mode
What it is	the equal share	the middle in order	the commonest value
Uses every value	yes	only their order	only the counts
A few huge values	pull it a long way	hardly move it	don’t move it
Works for words (colors, sizes)	no	only if they have an order	yes

• Use the mean when the total matters: the mean wage times the number of workers is the wage bill; Kathmandu’s mean monthly rain times 12 is the year’s rain. Weather offices worldwide describe a place’s climate by such means over 30 years, its climate normals.
• Use the median when a few values are far from the rest: incomes, house prices, Kathmandu’s monthly rain. In 2025, for example, the median household income in the United States was $87,460 and the mean $126,700 (the US Census Bureau’s figures): the mean is pulled up by the richest households, so official reports lead with the median.
• Use the mode for the most common choice: a shirt size to stock, a bus time most people want, a winner.

Question 1

Find the arithmetic mean of each.

Question 2

Find the median of each.

Question 3

Find the median of the data of Question 1.

Question 4

Find the arithmetic mean of the data of Question 2.

Question 5

Find the missing value in each.

Question 6

Find the mode of each.

Question 7

The median cuts ordered data into two halves. Cut each half again and the data is in four quarters. Health workers use cuts like these: a child’s height is checked against the World Health Organization’s growth standards, which give the heights at such cuts for every age. A child at the 25% mark is taller than about a quarter of healthy children of that age and shorter than three quarters.

Activity 1: a line in four parts

Activity 2: seven values

a. The smallest is 35. In ascending order: 35, 37, 40, 45, 46, 47, 48.

b. 35, 37, 40, 45, 46, 47, 48: the 2nd value, 37, has a quarter of the data before it; the 4th, 45, is the median; the 6th, 47, has three quarters of the data before it.

c. $25\% = \frac{25}{100} = \frac{1}{4}$, one part out of 4; $75\% = \frac{75}{100} = \frac{3}{4}$, three parts out of 4.

d. There are N = 7 items, so N + 1 = 8.

$$\begin{aligned} Q_1 &= \left(\frac{N + 1}{4}\right)\text{th item} = \left(\frac{8}{4}\right)\text{th item} \\ &= 2\text{nd item} = 37 \end{aligned}$$

$$\begin{aligned} Q_3 &= \left(\frac{3(N + 1)}{4}\right)\text{th item} = \left(\frac{3 \times 8}{4}\right)\text{th item} \\ &= 6\text{th item} = 47 \end{aligned}$$

So Q₁ = 37 and Q₃ = 47.

Activity 3: a discrete series

a. The weights are already in ascending order.

Weight (kg)	Frequency (f)	cf
20	6	6
22	8	14
24	9	23
27	7	30
28	5	35
30	5	40
31	3	43
	N = 43

b. Q₁ is the $\left(\frac{43 + 1}{4}\right)$th = 11th item; Q₃ is the $\left(\frac{3 \times 44}{4}\right)$th = 33rd item.

c. Items 7 to 14 weigh 22 kg, so Q₁ = 22 kg; items 31 to 35 weigh 28 kg, so Q₃ = 28 kg.

The quartiles are the three values that divide data, arranged in ascending order, into four equal parts. The first quartile Q₁ (the lower quartile) has 25% of the data at or below it; the second quartile Q₂ has 50% and is the median; the third quartile Q₃ (the upper quartile) has 75%. Their places are

$$Q_1 = \left(\frac{N + 1}{4}\right)\text{th item}$$

$$Q_3 = \left(\frac{3(N + 1)}{4}\right)\text{th item}$$

The middle half of the data lies between Q₁ and Q₃. In Activity 3, half the people weigh from 22 to 28 kg.

When the place is not a whole number, the quartile lies between two items. A place like 9.75th means the 9th item plus 0.75 of the step to the 10th:

$$\begin{aligned} 9.75\text{th item} = {} & 9\text{th item} \\ & + 0.75 \times (10\text{th item} - 9\text{th item}) \end{aligned}$$

The median’s 12.5th item in Example 5 of 18.2 was the same rule with 0.5.

Worked examples: quartiles

The opening asked how much rain a typical month gets in Kathmandu.

A typical month gets about 53 mm, the median; the mean, 125 mm, is the right number only for the year’s total.

In ascending order the 12 monthly normals are 6.4, 7.8, 15.2, 23.6, 38.0, 43.3, 62.9, 128.8, 207.8, 241.4, 342.7 and 384.1 mm. The median is the $\frac{12 + 1}{2} =$ 6.5th item, halfway between 43.3 and 62.9: 53.1 mm. The mean is $\frac{1502}{12} \approx$ 125.2 mm, pulled up by four monsoon months that together bring 1,176 mm; only 5 months get as much as the mean. There is no mode: all 12 values are different.

The quartiles say more. Q₁ is the 3.25th item, 15.2 + 0.25 × 8.4 = 17.3 mm, and Q₃ the 9.75th, 207.8 + 0.75 × 33.6 = 233.0 mm. So a quarter of the year’s months get under about 17 mm and a quarter over about 233 mm: Kathmandu has a dry season and a wet one, and no single “average month” describes either.

Grade 9’s final exam is set to CDC’s specification grid for Compulsory Mathematics, Grades 9–10. There, Statistics and Probability is 11 marks in two questions. In the grid’s model paper, question 14 gives the marks of 20 students as a discrete series:

Part	Level	Marks	What it asks
a.	knowledge	1	in which series the data is presented (Unit 17)
b.	understanding	1	a cumulative frequency table (Unit 17)
c.	application	2	the median
d.	higher ability	1	the mean of the marks above the median

Here is one of our own in the same shape.

Question 1

Find the first quartile and the third quartile of each.

Question 2

Find the first quartile Q₁ and the third quartile Q₃ of each.

Question 3

Question 4

Question 5

Find the first quartile and the third quartile of each.

Question 6

Find the first quartile Q₁ and the third quartile Q₃ of each.