S2, S3, S4StatisticsFoundation & HigherGrades 3–6

Frequency Tables

Q: What does 20 < t 25 mean?

Times greater than 20 and up to and including 25. A time of exactly 25 goes in this class; exactly 20 goes in the class below.

Q: Can I find the exact median from grouped data?

No — only the class containing it. At Higher, a cumulative frequency graph gives an estimate of the median.

Q: Why is the range not exact for grouped data?

Because the smallest and largest actual values are unknown; you only know the class limits.

A frequency table records how often each value, or each group of values, occurs. It saves writing out long lists — but it also changes how you find averages. With grouped data the exact values are lost, so the mean can only be $\textbf{estimated}$ using class midpoints.

45 min Video by Maths Genie AQA GCSE Maths

The big picture

A frequency of 6 next to the value 3 means "3, 3, 3, 3, 3, 3". So the total of all the values is found by multiplying each value by its frequency and adding: $\sum fx$ . The mean is then $\frac{\sum fx}{\sum f}$ .

For grouped data such as $10 < t \le 20$ , you do not know where each value lies within its class, so you assume they sit at the midpoint. That is why the answer is an estimate — and why the question always says "estimate".

Averages from Frequency Tables (incl. Estimating the Mean)Curated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

What you'll learn

Find the mean, median, mode and range from a frequency table
Find the modal class and the class containing the median for grouped data
Estimate the mean from a grouped frequency table
Complete and use two-way tables

Discrete frequency tables

Mode: the value with the highest frequency. Median: find the position $\frac{n + 1}{2}$ and count through the frequencies to reach it. Mean: add an $fx$ column, then divide $\sum fx$ by $\sum f$ .

Example 1The number of siblings of 20 students:

\begin{array}{c|c|c|c|c} \text{Siblings} & 0 & 1 & 2 & 3 \\ \hline \text{Frequency} & 4 & 9 & 5 & 2 \end{array}

Find the mode, median and mean.

1Mode: 1 sibling (frequency 9).

2Median: position

\frac{21}{2} = 10.5

. Running totals: 4, 13, … so the 10th and 11th values are both 1. Median

= 1

\sum fx = 0 + 9 + 10 + 6 = 25

. Mean

= 25 \div 20 = 1.25

Answer

Mode 1, median 1, mean 1.25

Your turnSame method, new numbers — try it before moving on.

Scores on a quiz: score 1 (frequency 3), score 2 (frequency 7), score 3 (frequency 6), score 4 (frequency 4). Find the mean score.

Your answer

Show hint

Total of scores $= 1 \times 3 + 2 \times 7 + 3 \times 6 + 4 \times 4$ .

Show worked solution

Solution

1 $\sum fx = 3 + 14 + 18 + 16 = 51$ .
2 $\sum f = 20$ , so the mean is $51 \div 20 = 2.55$ .

Answer: $2.55$

Tip — The mean is $\sum fx \div \sum f$ — dividing by the number of rows is the most common error.

Grouped frequency tables

The $\textbf{modal class}$ has the highest frequency. The class containing the median is found by counting cumulative frequencies to position $\frac{n}{2}$ (for large data sets).

To estimate the mean: find the midpoint of each class, multiply by the frequency, add, and divide by the total frequency.

Example 1Times (minutes) for 40 runners:

\begin{array}{c|c} \text{Time } t & \text{Frequency} \\ \hline 20 < t \le 25 & 6 \\ 25 < t \le 30 & 15 \\ 30 < t \le 35 & 12 \\ 35 < t \le 40 & 7 \end{array}

Estimate the mean time and state the modal class.

1Midpoints: 22.5, 27.5, 32.5, 37.5.

\sum fx = 135 + 412.5 + 390 + 262.5 = 1200

3Estimated mean

= 1200 \div 40 = 30

minutes. Modal class:

25 < t \le 30

Answer

Mean ≈ 30 minutes; modal class

25 < t \le 30

Your turnSame method, new numbers — try it before moving on.

Masses: $0 < m \le 10$ (frequency 5), $10 < m \le 20$ (frequency 8), $20 < m \le 30$ (frequency 7). Estimate the mean mass.

Your answer

Show worked solution

Solution

1Midpoints 5, 15, 25.
2 $\sum fx = 25 + 120 + 175 = 320$ ; $\sum f = 20$ .
3Estimated mean $= 320 \div 20 = 16$ .

Answer: $16$

Example 2For the runners, which class contains the median?

n = 40

, so the median is around the 20th–21st value.

2Cumulative frequencies: 6, 21, … — the 20th and 21st are in the second class.

Answer

25 < t \le 30

Why it is only an estimate

The estimate is only as good as the midpoint assumption. If most runners in $25 < t \le 30$ actually finished close to 26 minutes, the true mean would be lower than 30. That is why "explain why your answer is an estimate" is a common one-mark question: "the exact values are not known; midpoints were used".

Two-way tables

A two-way table shows two categories at once. Row and column totals let you fill in missing cells by subtraction.

Example 160 students chose French or Spanish. 34 are girls. 20 girls chose French. 38 students chose Spanish. How many boys chose Spanish?

1Boys:

60 - 34 = 26

. Girls choosing Spanish:

34 - 20 = 14

2Spanish total 38, so boys choosing Spanish:

38 - 14 = 24

Answer

24

Your turnSame method, new numbers — try it before moving on.

In a club of 45 members, 25 are adults. 18 members swim, and 11 of the swimmers are children. How many adults do not swim?

Your answer

Show worked solution

Solution

1Adults who swim: $18 - 11 = 7$ .
2Adults who do not swim: $25 - 7 = 18$ .

Answer: $18$

Think like an examiner

Add an $fx$ column to the table and show its total — this is the method mark.
Give the modal class as a class ( $25 < t \le 30$ ), not as a frequency (15).
For "estimate the mean", midpoints are essential; using class ends loses the method mark.

From frequency tables

\bar{x} = \dfrac{\sum fx}{\sum f}

Mean from a frequency table.

\text{midpoint} = \tfrac{\text{lower} + \text{upper}}{2}

Used to estimate the mean for grouped data.

\text{median position} = \tfrac{n + 1}{2}

Count through cumulative frequencies.

Watch out for these

Mean

= \sum fx \div

number of rows.

Divide by the total frequency

\sum f

Modal class given as the highest frequency, e.g. "15".

Give the class interval.

Using the upper class boundary instead of the midpoint.

Use

\frac{\text{lower} + \text{upper}}{2}

Median

=

middle row of the table.

Find the middle position by counting frequencies.

Stretch yourself

A frequency table has values 1, 2, 3 and 4 with frequencies 5, $x$ , 8 and 2. The mean is 2.2. Find $x$ .

Hint — Write $\sum fx$ and $\sum f$ in terms of $x$ , then set up $\frac{\sum fx}{\sum f} = 2.2$ .

Reveal solution

Solution

\sum fx = 5 + 2x + 24 + 8 = 37 + 2x

and

\sum f = 15 + x

. So

37 + 2x = 2.2(15 + x) = 33 + 2.2x

, giving

4 = 0.2x

and

x = 20

. Check:

\sum fx = 77

\sum f = 35

, and

77 \div 35 = 2.2

✓.

\blacksquare

Questions students ask

What does

20 < t \le 25

mean?

Times greater than 20 and up to and including 25. A time of exactly 25 goes in this class; exactly 20 goes in the class below.

Can I find the exact median from grouped data?

No — only the class containing it. At Higher, a cumulative frequency graph gives an estimate of the median.

Why is the range not exact for grouped data?

Because the smallest and largest actual values are unknown; you only know the class limits.

Key takeaways

Mean from a table: $\sum fx \div \sum f$ .
Mode: highest frequency; median: count to position $\frac{n + 1}{2}$ .
Grouped data: modal class, class containing the median, estimated mean.
Estimate the mean using class midpoints.
Two-way tables: use row and column totals to fill gaps.

Exam-style questions

Practise it the way it is examined

2 original questions · 7 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 33 marks

The number of pets owned by 30 people: $\begin{array}{c|c|c|c|c} \text{Pets} & 0 & 1 & 2 & 3 \\ \hline \text{Frequency} & 8 & 11 & 7 & 4 \end{array}$

Work out the mean number of pets.[3]
Show mark scheme
- M1 $\sum fx = 0 + 11 + 14 + 12 = 37$
- M1 $37 \div 30$
- A1 $1.23$ (1.233…)
Answer: $1.23$

Question 2Grade 54 marks

Heights of 50 plants (cm): $\begin{array}{c|c} \text{Height } h & \text{Frequency} \\ \hline 0 < h \le 10 & 7 \\ 10 < h \le 20 & 18 \\ 20 < h \le 30 & 16 \\ 30 < h \le 40 & 9 \end{array}$

(a)Work out an estimate for the mean height.[3]
Show mark scheme (a)
- M1Uses midpoints 5, 15, 25, 35
- M1 $\sum fx = 35 + 270 + 400 + 315 = 1020$
- A1 $20.4$ cm
Answer: $20.4$ cm
(b)Explain why your answer is an estimate.[1]
Show mark scheme (b)
- B1The exact heights are not known; midpoints were used

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Values 1, 2, 3 with frequencies 4, 9, 2. The mode is…
- $2$
- $9$
- $3$
- $1$
Show answer and explanation
Answer: $2$
Highest frequency.
Midpoint of $30 < x \le 40$ ?
- $35$
- $30$
- $40$
- $10$
Show answer and explanation
Answer: $35$
$(30 + 40) \div 2$ .
Total frequency for frequencies 6, 11, 3?
- $20$
- $3$
- $11$
- $66$
Show answer and explanation
Answer: $20$
Add the frequencies.
Values 0, 1, 2 with frequencies 5, 10, 5. Mean?
- $1$
- $6.67$
- $20$
- $10$
Show answer and explanation
Answer: $1$
$20 \div 20$ .
Values 1–4, frequencies 2, 3, 3, 2. Median?
- $2.5$
- $2$
- $3$
- $5$
Show answer and explanation
Answer: $2.5$
5th value 2, 6th value 3.
Grouped data gives an estimated mean because…
- midpoints are used
- the frequencies are wrong
- the median is unknown
- the mode is missing
Show answer and explanation
Answer: midpoints are used
Exact values are unknown.
Classes $0$ – $10$ (f 4) and $10$ – $20$ (f 6). Estimated mean?
- $11$
- $10$
- $12.5$
- $15$
Show answer and explanation
Answer: $11$
$(20 + 90) \div 10$ .
25 values in 3 classes with frequencies 8, 10, 7. Median is in…
- the 2nd class
- the 1st class
- the 3rd class
- cannot tell
Show answer and explanation
Answer: the 2nd class
The 13th value: cumulative 8, 18.

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