Every value counts once
Group A has 2 scores with mean 4. Group B has 6 scores with mean 8. The groups have different sizes, so their means should not count equally.
Their totals are 2 × 4 = 8 and 6 × 8 = 48. The combined mean is (8 + 48) ÷ (2 + 6) = 7. The larger group pulls it closer to 8.
Recover the group totals
Group A has 3 scores with mean 6. Group B has 2 scores with mean 11. Multiply each mean by its own count: 3 × 6 = 18 and 2 × 11 = 22.
Combine totals and counts: (18 + 22) ÷ (3 + 2) = 40 ÷ 5 = 8. Averaging 6 and 11 directly would treat the groups as equally large.
Recover a missing group average
An average hides a total. Multiply the combined average by the combined count, subtract the known group’s total, then divide by the missing group’s count.
Eight scores have a combined mean of 7. Group A has 2 scores with mean 4. The remaining 6 scores belong to group B. Find B’s mean.
8 × 7 = 56; 2 × 4 = 8
(56 − 8) ÷ 6 = 8
Check: 10 scores have a combined mean of 9. Group A has 4 scores with mean 6. What is the mean of the 6 scores in group B?
Check the answer
(10 × 9 − 4 × 6) ÷ 6 = 11
Group B has mean 11. Rebuild the overall total to check: the known group and the recovered group must account for all the scores.
(4 × 6 + 6 × 11) ÷ 10 = 9
Combine two unequal groups
Group A has 2 scores with mean 5. Group B has 6 scores with mean 9. Find the mean of all eight scores.
Recover the two group totals, add them and divide by the full count. The answer should lie between 5 and 9, closer to the larger group’s mean.
Try it
What is the combined mean for 2 scores with mean 5 and 6 scores with mean 9?
- 7
- 8
Show the answer
8
Yes. The group totals are 10 and 54. Divide their sum by 2 + 6 = 8 scores: 64 ÷ 8 = 8.
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Practice Ratios & averages