Averages from Frequency Tables

Find the mean, median, mode and total from a frequency table, and find a missing frequency when you know the mean.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024 · 56 real past-paper questions below

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What a Frequency Table Tells You

A frequency table is a short way to write a long list. Each row gives a value and how many times it appears.

Cousins	Frequency
0	1
1	7
2	5
3	4
4	3

The row "1 and 7" means 7 students each have 1 cousin. That is 7 people, not 1.

Key idea

Add the frequency column to get how many people there are: here $1 + 7 + 5 + 4 + 3 = 20$ students.

The Mean: Multiply, Add, Divide

The 3-mark question: "Work out the mean number of cousins." Each value counts as many times as its frequency.

Formula

mean = \frac{\sum f x}{\sum f}

Here $x$ is the value and $f$ is its frequency. The sign $\sum$ means "add them all up".

Worked example

1
$0 \times 1 + 1 \times 7 + 2 \times 5 + 3 \times 4 + 4 \times 3$ — Multiply each value by its frequency, row by row. The 7 students with 1 cousin have 7 cousins between them.
2
$= 0 + 7 + 10 + 12 + 12$ — Work out each product. Writing this line earns the first method mark.
3
$\sum f x = 41$ — Add the products. This is all the cousins of all the students.
4
$\sum f = 1 + 7 + 5 + 4 + 3 = 20$ — Add the frequency column. This is the number of students, not the 5 rows.
5
$\frac{41}{20} = 2.05$ — Share the cousins equally: total ÷ number of students. A mean does not have to be a whole number.

Rule

The 2-mark question "Work out the total number of cousins" stops at the coloured 41. Do not divide.

Common mistake

Dividing by 5, the number of rows, or averaging the values 0 to 4. Divide by the 20 students.

Key idea

Show the products line and the division. They earn 2 of the 3 marks, even if you slip on the adding.

Mode and Median From the Table

The 1-mark mode and the 2-mark median use the same table. Neither needs any multiplying.

Worked example

1
$mode = 1$ — The biggest frequency is 7. Its value is 1 cousin, so more students have 1 cousin than any other number.
2
$\frac{20 + 1}{2} = 10.5$ — There are 20 students. 10.5 means the middle is halfway between the 10th and 11th students.
3
$1$ — Running total after the 0 row: only the 1st student has 0 cousins. Not at the 10th yet.
4
$1 + 7 = 8$ — After the 1 row: the 2nd to 8th students have 1 cousin. Still short of the 10th.
5
$8 + 5 = 13$ — After the 2 row: the 9th to 13th students have 2 cousins. The 10th and 11th are in here.
6
$median = 2$ — Both middle students have 2 cousins, so the median is 2.

Common mistake

Giving 7 as the mode. 7 is how many times; the mode is the value, 1.

Key idea

A grouped table works the same way: the modal class is the class with the biggest frequency.

Key idea

Why is the mode the laziest average? In a frequency table it just finds the biggest frequency and points at the value. The mean has to multiply, add and divide, and it is still not finished until it checks the answer!

Median Between Two Values

Sometimes the two middle people have different values. Then the median is halfway between them. Here are the pets of 12 students.

Pets	Frequency
0	3
1	3
2	2
3	3
4	1

Formula

median position = \frac{n + 1}{2}

$n$ is the total frequency: how many people there are. The answer tells you which person is in the middle, not their value.

Worked example

1
$n = 3 + 3 + 2 + 3 + 1 = 12$ — Add the frequency column to count the students.
2
$\frac{12 + 1}{2} = 6.5$ — The middle is halfway between the 6th and 7th students.
3
$3$ — Running total after the 0 row: the 1st to 3rd students have 0 pets. Not at the 6th yet.
4
$3 + 3 = 6$ — After the 1 row: the 4th to 6th students have 1 pet. So the 6th student has 1.
5
$6 + 2 = 8$ — After the 2 row: the 7th and 8th students have 2 pets. So the 7th student has 2.
6
$\frac{1 + 2}{2} = 1.5$ — The two middle values differ, so add them and halve. The median is 1.5 pets.

Common mistake

Giving 6.5 as the median. 6.5 is the position of the middle student; the median is their value, 1.5.

Range From the Table

“Work out the range” is a 1-mark question. Use the value column only, and skip any value nobody has.

Formula

range = largest value - smallest value

Here a value means an answer someone actually gave: its frequency is not 0. These are the texts sent by 20 students in one hour.

Texts	Frequency
1	0
2	5
3	8
4	4
5	2
6	1

Worked example

1
$1 has frequency 0$ — Nobody sent 1 text, so 1 is not in the data. Skip it.
2
$smallest = 2$ — The first value with a frequency above 0 is 2 texts.
3
$largest = 6$ — 6 has frequency 1, so one student sent 6 texts. That is the largest value.
4
$range = 6 - 2 = 4$ — Largest take away smallest. The range is 4 texts.

Common mistake

Using the frequencies: 8 − 0 = 8. The range is about the values, not how often they appear.

Missing Frequency From the Mean

The 4-mark question: one frequency is $x$ , and you are told the mean. A team scored these goals per game, and the mean is 1.5.

Goals	Frequency
0	4
1	9
2	$x$
3	3

Worked example

1
$0 \times 4 + 1 \times 9 + 2 \times x + 3 \times 3$ — Multiply each value by its frequency, as for any mean. The unknown row gives $2 \times x$ .
2
$\sum f x = 18 + 2 x$ — 0 + 9 + 9 = 18. The $2 x$ stays as it is, because $x$ is unknown. These are the goals.
3
$\sum f = 4 + 9 + x + 3 = 16 + x$ — Add every frequency, including the $x$ games. These are the games.
4
$\frac{18 + 2 x}{16 + x} = 1.5$ — Mean = total ÷ total frequency, and the mean is 1.5. This equation earns the method marks.
5
$18 + 2 x = 1.5 (16 + x)$ — Multiply both sides by $16 + x$ to clear the fraction.
6
$18 + 2 x = 24 + 1.5 x$ — Expand the bracket: 1.5 × 16 = 24 and 1.5 × $x$ = $1.5 x$ .
7
$18 + 0.5 x = 24$ — Take $1.5 x$ from both sides, so the $x$ terms are only on the left.
8
$0.5 x = 6$ — Take 18 from both sides.
9
$x = 12$ — Divide both sides by 0.5. $x$ counts games, so it must be a whole number, and 12 is.

Common mistake

Dividing by 16. The $x$ games count too, so the total frequency is $16 + x$ .

Key idea

Check by putting $x = 12$ back: $\frac{18 + 24}{16 + 12} = \frac{42}{28} = 1.5$ . It works. If you get a negative or a fraction, look for a slip.

Which Method? Read the Question

The table is the same every time. The question word tells you which column to use.

You see	You do
Mode	value with the biggest frequency
Median	running total to the middle
Range	largest value − smallest value (skip frequency 0)
Mean	add value × frequency, ÷ total frequency
Total number of ...	add value × frequency, stop
A bar chart of values	bar height = frequency, then the same methods
Mean given, one frequency is $x$	total ÷ frequency = mean, solve
Just a list of numbers	Mean, Median, Mode and Range
Classes like $10 < t \leq 20$	Estimating the Mean for Grouped Data

Key idea

Your mean must lie between the smallest and biggest value. A mean of 8.2 cousins from values 0 to 4 is wrong.

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