Lesson 33 of 100 | Quantitative Aptitude / Averages / औसत
Combined and Weighted Averages
Learning outcome
Combine unequal groups correctly, calculate weighted scores and recover a group-size ratio from its combined mean.
Meaning and method
A group's total = count × mean. To combine nonempty groups, add their totals and divide by the combined count.
Simply averaging the means gives equal influence to both groups, regardless of their sizes. Equal positive counts make this shortcut valid; for unequal counts, use their actual weights instead.
A weighted mean also describes values assigned different importance. Multiply each value by its weight, add these products, then divide by the sum of the weights. Values must use a common scale. Group-size weights count observations; assessment weights specify their relative importance.
Weights are nonnegative, and their total must be positive. Omit zero-weight entries. An empty group's mean is undefined, so do not invent a mean for it. If all counts or weights are zero, no average is defined.
Multiplying every weight by the same positive factor changes numerator and denominator equally, leaving the mean unchanged. Thus a weight ratio is enough. With positive weights, the result lies between the smallest and largest included values.
Worked examples
Example 1 — Combine groups. Eight learners average 35 pages each; twelve average 45 pages each. Total pages = 8 × 35 + 12 × 45 = 280 + 540 = 820. Combined mean = 820 ÷ 20 = 41 pages. The larger group pulls the result closer to 45; simply averaging the means would give 40.
Example 2 — Apply importance weights. Three tests, each scored out of 100, have scores 64, 76 and 88 with weights 2:3:5. Weighted score = (2 × 64 + 3 × 76 + 5 × 88) ÷ (2 + 3 + 5) = 796 ÷ 10 = 79 3/5. Divide by total weight, not the number of tests.
Example 3 — Recover a count ratio. Two groups have means 24 and 39; together their mean is 33. The first mean is 33 - 24 = 9 below the combined mean; the second is 39 - 33 = 6 above. For counts a and b, total deficit balances total excess: 9 × a = 6 × b. Thus a:b = 6:9 = 2:3.
Common mistakes
Do not confuse a group mean with its total. Keep weights in the same order as their values. More observations give a group greater influence; a higher mean alone gives no extra weight.
Practice questions
- Combine 6 learners averaging 21 marks with 9 learners averaging 31 marks.
- Two group means are 18 and 32; their counts are in ratio 3:4. Find the combined mean.
- Scores 50, 65 and 80 have weights 1:2:3. Find the weighted mean.
- Groups with means 16 and 28 have combined mean 25. Find their count ratio in that order.
Worked answers
- Total = 6 × 21 + 9 × 31 = 126 + 279 = 405. Count = 15, so mean = 405 ÷ 15 = 27 marks.
- Use the ratio as weights: (3 × 18 + 4 × 32) ÷ 7 = 182 ÷ 7 = 26.
- Weighted total = 50 + 130 + 240 = 420. Total weight = 6, so mean = 420 ÷ 6 = 70.
- The distances from 25 are 9 below and 3 above. Balancing gives 9 × a = 3 × b, so a:b = 3:9 = 1:3.