Estimated means and modal classes for grouped data
Use class midpoints when exact observations are unavailable and label the resulting mean as an estimate.
Extended only. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Find a weighted mean.
- Read continuous class boundaries.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What represents each grouped class in an estimated mean?
What you will learn
- Calculate an estimated grouped mean.
- Identify a modal class from grouped frequencies.
Midpoints stand in for hidden observations
For classes 0 ≤ x < 10, 10 ≤ x < 20 and 20 ≤ x < 30 with frequencies 3,5,2, use midpoints 5,15,25. The weighted sum is 3×5 + 5×15 + 2×25 = 140. Ten observations give estimated mean fourteen.
This is an estimate because the exact positions inside each class are unknown. Multiplying every class endpoint by frequency would make an unjustified extreme assumption. The midpoint is a representative value, not proof that every observation equals it.
A modal class is an interval, not an exact mode
For equal-width classes above, the class with largest frequency five is 10 ≤ x < 20. Report that interval as the modal class, rather than saying the exact mode is fifteen. Grouping may hide the original most frequent value.
A grouped table generally cannot produce the exact original median or range. Further graphical estimates use class boundaries and cumulative frequency. Histograms need special care with unequal widths: use frequency density for bar heights, so the tallest density bar need not have the largest frequency.
Pause and explain
Midpoint total is 140 for 10 observations. What is the estimated mean?
Worked example
Classes 0–10, 10–20, 20–30 have frequencies 3,5,2. Estimate the mean.
Show the worked solution
- Midpoints are five, fifteen and twenty-five.
- The weighted midpoint total is 15+75+50 = 140.
- Divide by total frequency ten to get estimated mean fourteen.
Answer Estimated mean 14.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Classes 0–10 and 10–20 have frequencies 2 and 3. Estimate mean.
Give me a hint
Use midpoints five and fifteen.
Compare my reasoning
- Weighted sum is 2×5 + 3×15 = 55.
- Total frequency is five.
- Estimated mean is eleven.
Estimated mean 11.
Look for these in your work
- Used class midpoints.
- Labelled the answer as estimated.
Equal-width classes 0–5,5–10,10–15 have frequencies 4,8,3. State the modal class.
Give me a hint
Find the largest class frequency.
Compare my reasoning
- The largest frequency is eight.
- It belongs to the interval 5 ≤ x < 10.
- The exact most frequent value inside that interval is unknown.
5 ≤ x < 10.
Look for these in your work
- Reported an interval.
- Avoided claiming an exact mode.
Explain why two raw data sets can have the same grouped table but different exact means.
Give me a hint
Values can move within their classes without changing counts.
Compare my reasoning
- One data set may place values near each lower boundary.
- Another can place them near upper boundaries while preserving all frequencies.
- Their totals differ, so the grouped midpoint mean cannot be exact for both.
Class counts hide positions within intervals, so exact means can differ.
Look for these in your work
- Identified lost information.
- Explained why midpoint estimates are limited.
Common mistakes
- Forgetting midpoint weighting.
- Calling the estimate exact.
- Giving a midpoint as a known exact mode.
What can you explain now?
Can a modal class tell you the exact mode?
Compare with the explanation
No.
Grouping hides the individual values and their exact repetition counts.
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Make it stick
Tomorrow, calculate a grouped estimate and explain one piece of information lost by grouping.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.