Combine two datasets by restoring and adding moments
Weight means by counts, reconstruct second totals and include the spread between group centres.
Section 5.1 · Representation of data
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.1. Label variables, units, class boundaries and diagram keys. In these raw-data quartile exercises, use the median of each ordered half, excluding the overall median when the number of observations is odd; follow any convention supplied in an examination question. Cumulative-frequency quartiles use ranks n/4 and 3n/4, with a median rank n/2. Grouped-data interpolation and midpoint moments are estimates, not recovered raw measurements. Histogram area represents frequency, with height proportional to frequency density. Use descriptive variance with divisor n, not the sample-estimator divisor n − 1. Coded and combined totals must refer to the same units and compatible coding before addition. Retain unrounded working and distinguish variance units from standard-deviation units. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Calculate descriptive first and second moments.
- Decode any shifted or scaled totals into common units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Groups have counts 4 and 6 with means 10 and 20. Their combined mean is…
What you will learn
- Combine up to two compatible datasets using totals or means and standard deviations.
- Explain why averaging group standard deviations is invalid.
Add counts and totals, not unweighted summaries
Group A has n = 4, mean 10 and descriptive standard deviation 2. Its sum is 40 and squared sum is 4(2² + 10²) = 416. B has n = 6, mean 20 and standard deviation 3, so sum 120 and squared sum 6(3² + 20²) = 2454. Combined n = 10, sum 160 and squared sum 2870; mean is 16 and variance 287 − 16² = 31, giving standard deviation √31 ≈ 5.568.
The combined mean is (nA meanA + nB meanB)/(nA + nB). The combined squared total uses n(variance + mean²) for each group. Do not use the unweighted mean 15 when counts differ. Standard deviations cannot simply be averaged: dispersion between group centres also contributes to the combined spread.
Compatibility and grouped estimates matter
If two groups have equal mean 5, counts 3 and 2 and descriptive standard deviations 1 and 2, the combined variance is (3(1²) + 2(2²))/5 = 11/5, giving standard deviation √(11/5). The between-mean contribution vanishes only because the means are equal. Even here, averaging the standard deviations is wrong; variance, not standard deviation, enters the second moment.
Use the same measurement units, definition of descriptive variance and coding before combining. If moments came from class midpoints, the combined result remains an estimate. Counts cannot be added to combine incompatible populations without explaining the purpose, and pooled descriptive summaries do not by themselves establish population-level inference or eliminate differences in measurement conditions.
Pause and explain
For n = 4, mean 10 and descriptive SD 2, what is Σx²?
Worked example
Combine A (n 4, mean 10, descriptive SD 2) and B (n 6, mean 20, descriptive SD 3) in the same units.
Show the worked solution
- Recover A totals 40, 416 and B totals 120, 2454 from n mean and n(SD² + mean²).
- Add to n = 10, Σx = 160, Σx² = 2870; combined mean = 16.
- Combined variance = 287 − 256 = 31; SD = √31 ≈ 5.568.
Answer Combined mean 16 and descriptive standard deviation √31 ≈ 5.568.
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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.
Combine A (n 4, mean 10, descriptive SD 2) and B (n 6, mean 20, descriptive SD 3) in the same units.
Give me a hint
Restore each first and second total before adding.
Compare my reasoning
- A totals are 40 and 416; B totals are 120 and 2454.
- Combined n = 10, sum 160 and squared sum 2870, giving mean 16.
- Variance = 2870/10 − 16² = 31 and SD = √31 ≈ 5.568.
Combined n 10, mean 16, variance 31 and SD √31 ≈ 5.568.
Look for these in your work
- Weight all observations through counts.
- Include squared means when reconstructing moments.
Two groups share mean 5 and have (n, descriptive SD) = (3,1) and (2,2). Find combined mean and SD.
Give me a hint
Equal means remove the between-centre term, but variances still need count weights.
Compare my reasoning
- Their first totals are 15 and 10; combined mean is 25/5 = 5.
- Squared totals are 3(1 + 25) = 78 and 2(4 + 25) = 58.
- Combined variance = 136/5 − 25 = 11/5; SD = √(11/5) ≈ 1.483.
Combined mean 5; variance 11/5; SD √(11/5) ≈ 1.483.
Look for these in your work
- Weight variances by group counts when means match.
- Do not average standard deviations.
A has three observations coded y = x − 10 with Σy = 3 and Σy² = 5. B has two observations coded z = x − 20 with Σz = 2 and Σz² = 4. Combine in original x units.
Give me a hint
The different origins must be undone before their totals are added.
Compare my reasoning
- A gives Σx = 33 and Σx² = 5 + 20(3) + 100(3) = 365.
- B gives Σx = 42 and Σx² = 4 + 40(2) + 400(2) = 884.
- Combined n = 5, sum 75 and squared sum 1249; mean 15, variance 1249/5 − 225 = 124/5, SD √(124/5) ≈ 4.980.
Combined mean 15; variance 124/5; SD √(124/5) ≈ 4.980 in original units.
Look for these in your work
- Decode each group with its own origin.
- Add compatible original moments rather than different coded totals.
Common mistakes
- Weight group means by their counts.
- Recover and add squared totals, not standard deviations.
- Combine units and coding consistently; preserve grouped-estimate status.
What can you explain now?
Why can the combined standard deviation exceed both group standard deviations?
Compare with the explanation
The difference between group means adds spread in the pooled data.
Within-group spreads alone omit the separation of the groups’ centres.
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Make it stick
Restore n, sum and squared sum in common units, add them, and calculate the pooled descriptive mean and spread.
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.