How this instrument works
A percentage on its own is silent about how many things it was measured over, and averaging two percentages without accounting for that hides a real difference in weight. This formula restores it: multiply each percentage by the sample size that produced it, add the two products, and divide by the combined sample size. What looks like an ordinary weighted mean is secretly an exact recovery of the pooled counts — multiplying Percentage 1 by Sample size 1 yields a number proportional to how many items actually passed, without ever writing the divide-by-100 that would normally turn a percentage into a count.
The proof fits in one line. The true count in group one is (p₁ ⁄ 100) × n₁, the true count in group two is (p₂ ⁄ 100) × n₂, and the combined percentage is 100 times their sum divided by n₁ + n₂. Both factors of 100 meet in that last step and cancel, leaving (p₁n₁ + p₂n₂) ⁄ (n₁ + n₂) standing on its own — a formula that never mentions 100 yet is exactly equivalent to counting successes and dividing by the total. Set n₁ equal to n₂ and the weighting disappears, collapsing the result to the ordinary average of the two percentages; equal group size is the one case where a naive mean happens to already be correct.
Push the sizes far apart and the behaviour turns genuinely counter-intuitive: shrink Sample size 1 toward zero while holding Sample size 2 fixed, and the combined figure drifts all the way to Percentage 2, no matter how far apart the two rates started. A single well-measured batch of ten thousand outweighs a shaky batch of five almost entirely — the right behaviour for a pooled estimate, but the wrong instinct if percentages were ever learned as two numbers that simply split the difference.
- Enter the first group's rate in Percentage 1 and how many items it covers in Sample size 1.
- Enter the second group's rate in Percentage 2 and how many items it covers in Sample size 2.
- Read Weighted average percentage — the single pooled figure across both groups combined.
- Swap either pair of numbers to see how a size difference pulls the pooled result toward the larger group.
Worked example — two batches, 20 and 30 units
A quality-control line inspects two batches on one shift. Batch one, Percentage 1 = 80 with Sample size 1 = 20, passes 80% of its 20 units. Batch two, Percentage 2 = 90 with Sample size 2 = 30, passes 90% of its 30. The instrument combines them as (80 × 20 + 90 × 30) ⁄ (20 + 30) = (1600 + 2700) ⁄ 50 = 4300 ⁄ 50 = 86 — the Weighted average percentage for the full shift is 86%.
Simply averaging 80 and 90 gives 85, and that figure is wrong for the shift as a whole: it silently treats both batches as if they held the same 25 units. The true count behind the numbers is 16 passes out of 20 in the first batch and 27 out of 30 in the second, 43 passes out of 50 units in total, and 43 ⁄ 50 is exactly 86% — the same answer the formula reaches without ever dividing by 100 along the way.
Questions
Why is the answer 86% and not the plain average of 85%?
Because the plain average of 80 and 90 assumes the two batches are the same size, and here they are not. Weighting by Sample size 1 (20) and Sample size 2 (30) pulls the result toward the larger batch's 90%, landing at 86% — the same figure you would get by counting 16 passes out of 20 and 27 out of 30, then dividing 43 by 50.
What happens when both sample sizes are equal?
The formula collapses to the plain average, (p₁ + p₂) ⁄ 2. When n₁ = n₂, the weight each percentage carries is identical, so the count-based pooling and the simple mean produce the same figure. Equal weight is the one condition under which averaging two percentages directly happens to already be correct.
Why does the formula work without dividing p₁ and p₂ by 100 first?
Because that division cancels out algebraically. The true pass count in group one is (p₁ ⁄ 100) × n₁, and the pooled percentage is 100 times the total count over the total size — the two factors of 100 meet and disappear, leaving (p₁n₁ + p₂n₂) ⁄ (n₁ + n₂) as an exact shortcut rather than an approximation.
Can this handle more than two groups?
Not directly — this sheet is built for two. The same principle extends to any number of groups: multiply each percentage by its sample size, add every product, then divide by the sum of every sample size. Three or more groups need that longer sum worked out by hand or in a spreadsheet.
Is this the same kind of weighting used for exam grades or budget shares?
The arithmetic is identical but the weights mean something different. An exam grade weights scores by an assigned importance, say 70% coursework and 30% final, chosen ahead of time. Here the weights are the actual sample sizes n₁ and n₂ — how many items each percentage was measured over — so the pooling reflects real counts, not assigned importance.
What if one sample size is zero?
The instrument reports the other group's percentage untouched. With n₁ = 0, every term involving Percentage 1 drops out and avg equals Percentage 2 exactly, because a group with zero members contributes zero count and zero weight to the pool. The one case the sheet blocks is both sample sizes at zero together, since that divides by zero.