Maths · Suggested Grade 10
Class 10 Mean of Grouped Data: Midpoints & Three Methods
If one interval contains eight students and another contains two, their midpoints should not receive equal weight. A grouped mean uses frequency—and usually estimates the lost individual values.
Original explanations and practice by BananaLearn · Published 31 October 2026
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What grouping keeps and what it loses
A grouped frequency table gives the number of observations in intervals. It does not normally preserve every exact observation. We represent each interval by its midpoint x = (lower boundary + upper boundary)/2 and weight that midpoint by the class frequency f. The estimated mean is Σfx / Σf.
Read intervals consistently. Here 0–10 means 0 ≤ value < 10, then 10–20 includes 10 and excludes 20. An observation at 10 belongs to the second class, not both. The examples use continuous intervals or explicitly stated boundaries. Do not silently mix inclusive whole-number classes with continuous ones.
Worked example: the direct method
Suppose the intervals 0–10, 10–20 and 20–30 have frequencies 2, 5 and 3. Their midpoints are 5, 15 and 25. Multiply to obtain fx values 10, 75 and 75. The total frequency is 10 and the total fx is 160. The estimated mean is 160 ÷ 10 = 16.
Simply averaging 5, 15 and 25 gives 15, which ignores that the classes contain different numbers of observations. The midpoint model is like listing two 5s, five 15s and three 25s. It weights each of the ten represented observations equally, not each of the three intervals equally.
Why the answer is an estimate
The two observations in 0–10 could be 1 and 2 or 8 and 9, while the grouped table remains unchanged. Their exact contribution to the raw-data mean would differ. Midpoints replace that unknown within-class detail, so write estimated mean when the context requires it.
For some specially balanced raw datasets the midpoint calculation can happen to equal the exact mean. That coincidence does not make it exact for every possible dataset with the same table. Avoid giving more decimal places than the source data and question justify. A precise-looking decimal is not proof of precise measurement.
Worked example: use an assumed mean
Choose A = 15, one central midpoint. Deviations d = x − A are −10, 0 and 10. Multiply by frequencies 2, 5 and 3 to get fd = −20, 0 and 30. Their sum is 10. The mean estimate is A + Σfd/N = 15 + 10/10 = 16.
This method is the same calculation rearranged. Since x = A + d, Σfx = AΣf + Σfd. Divide by N = Σf to obtain A + Σfd/N. A can be any convenient value; it does not have to equal the final mean. Choosing a nearby central value usually makes the arithmetic simpler.
Worked example: step deviation for equal-width classes
The intervals above have common width h = 10. Set u = (x − A)/h, producing −1, 0 and 1. The fu values are −2, 0 and 3, totalling 1. Reconstruct the estimate as A + hΣfu/N = 15 + 10 × 1/10 = 16.
The width h restores the scale you divided out. Forgetting it would give 15.1, a different answer. For the usual school step-deviation table, equal-width classes make consecutive midpoint codes convenient. The direct method remains straightforward when class widths differ; do not invent a single common width for unequal intervals.
Worked example: compare two distributions
A second group has the same intervals but frequencies 3, 5 and 2. Its midpoint products are 15, 75 and 50, totalling 140 for ten observations. Its estimated mean is 14. Compared with the first table, one observation has moved from the highest class to the lowest class in the midpoint model.
That reduces the weighted total by 25 − 5 = 20, so the mean estimate falls by 20/10 = 2. This reasoning checks the arithmetic without redoing every multiplication. It describes the modelled change; the exact change in raw observations is not known from interval membership alone.
A table-checking routine before you submit
Check that the frequencies add to N, each midpoint lies inside its interval, and the fx column multiplies the correct row. A weighted average with non-negative frequencies must lie between the smallest and largest represented midpoints. An answer outside that range signals an arithmetic or denominator error.
Use one method completely, then verify with another if the question or revision task calls for it. Do not add the midpoints and divide by the total frequency; that mixes an unweighted numerator with a weighted denominator. In an explanation, state what f, x, A, h and N mean before using their symbols.
Complete the three-method comparison
Copy the record into a notebook or use it to organise your own answer. Blank spaces are for your work.
| Interval | f | x | fx | d = x − 15 | fd | u = d/10 | fu |
|---|---|---|---|---|---|---|---|
| 0–10 | 2 | 5 | ___ | ___ | ___ | ___ | ___ |
| 10–20 | 5 | 15 | ___ | ___ | ___ | ___ | ___ |
| 20–30 | 3 | 25 | ___ | ___ | ___ | ___ | ___ |
| Totals | 10 | — | ___ | — | ___ | — | ___ |
10 questions with explained answers
Try the questions before opening the answers. Explain the reason, then try a changed example to check your understanding.
Find the midpoint of 20–30.
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25
(20 + 30) ÷ 2.Frequencies are 2, 5, 3. Find N.
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10
Add the frequencies, not the number of intervals.For midpoints 5,15,25 and frequencies 2,5,3, find Σfx.
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160
10 + 75 + 75 = 160.Estimate the mean from that table.
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16
160 ÷ 10.With A = 15, find the deviations for those midpoints.
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−10, 0, 10
Subtract 15 from each midpoint.For the same table, find Σfd.
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10
2 × (−10) + 5 × 0 + 3 × 10 = 10.With h = 10, find Σfu.
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1
The codes are −1, 0, 1; weighted sum −2 + 0 + 3.Estimate the mean when the frequencies become 3,5,2.
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14
Σfx = 140 and N = 10.Why is the grouped midpoint mean usually an estimate?
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Exact values inside the intervals are not known
Different raw datasets can produce the same class frequencies.A calculation gives mean 31 using only midpoints 5,15,25 and positive frequencies. Is it possible?
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No
A positive-weight average must lie between 5 and 25.
Keep the ideas connected
- Review averages of ungrouped values
- Read grouped intervals and frequency
- Connect Class 10 revision topics
- Explore Class 10 Maths support
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