Equal-size groups
4 students average 72, and another 4 students average 84.
(72 + 84) ÷ 2 = 78
(4×72 + 4×84) ÷ 8 = 78
Safe: both averages represent 4 students.
Chapter 3 · Lesson 3.2
Combine Groups Correctly: Weighted Averages
正确合并小组：加权平均数
Learn why a larger group must have a larger influence, when it is safe to average two averages, and how totals keep every member counted fairly.
A group average speaks once for every member of its group.
By the end of this lesson, I can…
- explain why a group’s size is its weight;
- combine group totals and group counts correctly;
- decide when simply averaging two averages is safe;
- use ratios as weights when exact group sizes are not needed;
- work backward to find an unknown group size.
Combined average = combined total ÷ combined count
Build each group’s total first: average × number of members. For this combined-average formula, the groups must not overlap: count every member once. Use the same measurement units and exact means unless rounding is requested. The overlapping-group challenge later needs a separate correction.
Mission 1
Not completed
Every member has one vote in the combined average
Group A has only 2 students with an average of 90. Group B has 8 students with an average of 70. The larger group must influence the combined average more.
Group A
Average 90Hidden total: 2 × 90 = 180 points
Group B
Average 70Hidden total: 8 × 70 = 560 points
Count every student
74
(180 + 560) ÷ (2 + 8) = 740 ÷ 10 = 74
Where should the combined average land?
7090
74
Group B
Group A
Key idea: The two averages are not two equal-sized pieces. One represents 2 students; the other represents 8 students.
Hint
Turn each average back into a total. Add both totals, then divide by all 10 students.
Mission 2
Not completed
When is it safe to average two averages?
A simple average of group averages is guaranteed to work when the groups have the same number of members. Then each group really does have the same weight.
Unequal-size groups
2 students average 90, and 8 students average 70.
(90 + 70) ÷ 2 = 80
(2×90 + 8×70) ÷ 10 = 74
Unsafe: the two averages represent different numbers of students.
Reliable rule: When group sizes differ, do not average the averages. Combine totals and counts instead. Equal group sizes are a guaranteed safe case. If the two means are identical, that same mean is also the combined mean, even with unequal group sizes.
Hint
Ask whether each average stands for the same number of people. If yes, the weights match.
Mission 3
Not completed
Build the weighted-average formula from totals
“Weighted average” is not a mysterious new rule. It is the ordinary average formula after each group average has been turned back into its group total.
Group total = group average × group count
Combined average = (total A + total B) ÷ (count A + count B)
= (average A × count A + average B × count B) ÷ total count
Group A
Total A: 492
Group B
Total B: 288
Combined total780
Combined count10
Weighted average78
Simple mean of averages77
Difference1
Change the four machine inputs and watch the result. When counts are equal, the weighted and simple results match. When counts differ, they usually do not.
Fixed practice case: Group A has 6 students averaging 82. Group B has 4 students averaging 72. Answer these questions using this case, even if you have changed the exploration machine above.
Mission 4
Not completed
Build three equal-average groups from 1 through 9
The numbers 1 through 9 have a total of 45. Place each number exactly once into three groups of three so that all three groups have the same average. Order within a group does not matter.
1 + 2 + ··· + 9 = 45
45 ÷ 3 groups = 15 per group
15 ÷ 3 numbers = average 5
Group 1
Sum —Average —
Group 2
Sum —Average —
Group 3
Sum —Average —
Hint
Try pairing a small number with a large number. Every group must total 15, and every number from 1 to 9 must be used exactly once.
Does every group need three numbers?
The worked example only requires equal averages. Try {5}, {1, 9}, and {2, 3, 4, 6, 7, 8}. Their sizes are 1, 2, and 6, but every mean is 5.
If each group has mean m, its total is its count × m. Combining all nine numbers gives 9 × m = 45. Thus m = 5, and the three means sum to 15, even when the group sizes differ.
Mission 5
Not completed
The 25-and-15 trap
A class has 40 students. The top 25 students average 10 points more than the bottom 15. A student averages the two subgroup averages equally. How far below the true class average is that answer?
Top group
25 students · average 80Concrete model total: 25 × 80 = 2000
Bottom group
15 students · average 70Concrete model total: 15 × 70 = 1050
True class average
76.25
(2000 + 1050) ÷ 40 = 76.25
The larger top group pulls the class average upward.
7080
76.25
Bottom
Top
Why the error is always 1.25 points: The wrong method lowers each of 25 top students by 5 points but raises each of 15 bottom students by 5. The total falls by 25×5 − 15×5 = 50 points, so the class average falls by 50 ÷ 40 = 1.25.
Mission 6
Not completed
Two classes, one known combined average
Class 1 has 52 students and Class 2 has 48. Together, all 100 students average 78. Class 2’s average is 5 points higher than Class 1’s. Find both class averages without guessing.
52Class 1 students
48Class 2 students
1. Whole total
100 × 78 = 7800
2. Extra carried by Class 2
48 × 5 = 240
3. Remove the extra
7800 − 240 = 7560
4. Baseline average
7560 ÷ 100 = 75.6
Class 1 average = 75.6Class 2 average = 75.6 + 5 = 80.6
Mission 7
Not completed
Use a ratio when only relative group sizes matter
A class has half as many girls as boys. We are comparing the mean body mass, measured in kilograms. Boys average 41 kg and girls average 35 kg. We do not need the exact class size—the ratio 2 boys : 1 girl is enough. Use a representative model of two boys and one girl, each standing for the same number of actual students. The model total is not the whole class’s actual total mass.
Boys: 2 equal-size units
41 kgone boy-unit
41 kgone boy-unit
Girls: 1 unit
35 kgone girl-unit
(2×41 + 1×35) ÷ (2+1)117 ÷ 3 = 39 kg
Ratio weights: Multiplying every group size by the same number does not change the combined average. The ratio alone preserves the weights.
Hint
Use three equal-size units: 41, 41, and 35. Then find their average.
Mission 8
Not completed
Work backward to find a missing group size
A school has 100 competitors with an overall average of 63. Boys average 60 and girls average 70. How many girls and boys are there?
The class average is 3 points above the boys’ average.
60
boys’ average
63
whole class
70
girls’ average
Whole-class total100 × 63 = 6300
All-boy baseline100 × 60 = 6000
Extra above baseline6300 − 6000 = 300
Each girl accounts for, on average,70 − 60 = 10 extra points
Girls300 ÷ 10 = 30; therefore boys = 70
Distance idea: The overall average is 3 of the 10 points from 60 to 70. Therefore girls make up 3/10 of the 100 students: 30 students.
Mission 9
Not completed
Weighted-average workshop
For every problem, identify the group average and the weight it represents. Build totals before combining or working backward.
1Equal groups
Two classes each have 30 students. Their averages are 72 and 84.
2Unequal groups
20 students average 80 and 30 students average 70.
3Notebook distribution
A set of notebooks would give each girl 15 if only girls received them, or each boy 10 if only boys received them. Share all the notebooks equally among all boys and girls in the class. Each notebook costs ¥0.50, and each student pays for their own share.
4Ratio weights
There are twice as many boys as girls. Boys average 44 kg and girls 38 kg.
5Unequal-group error
25 students average 90 and 15 students average 80.
6Prize-group transfer
There were 10 first-prize and 20 second-prize students. Moving the 4 lowest-scoring first-prize students into second prize, without changing any scores, raises the first-prize average by 3 and the second-prize average by 1.
7Find a group count
100 students average 68. Group A averages 65 and Group B averages 75.
8Choose the guaranteed rule
When is it guaranteed that the simple mean of two group averages equals the combined average?
Hints for Questions 3, 6, and 7
3: If there are N notebooks, girls = N/15 and boys = N/10, so the whole class has N/6 students.
6: Let the original averages be A and B. The moved group’s total is both 10A−6(A+3) and 24(B+1)−20B.
7: Begin with all 100 at 65, then count the 10-point extras needed to reach the actual total.
Worked solution: notebook distribution
- Use a small model of 30 notebooks. They would serve 30 ÷ 15 = 2 girls or 30 ÷ 10 = 3 boys.
- Sharing among all 5 children gives 30 ÷ 5 = 6 notebooks each.
- Each pays 6 × 0.50 = 3 yuan. Scaling the model up changes the class size and notebook count together, so the per-child amount stays the same.
Worked solution: moving the prize group
- After four students leave the first group, the six remaining students average 3 more than before. Their total excess above the old average is 6 × 3 = 18. The four who left must balance this: their mean is 18 ÷ 4 = 4.5 below the old first-group mean.
- The second group grows from 20 to 24 and its mean rises by 1. The moved students bring 24 × 1 = 24 points above the old second-group baseline: 24 ÷ 4 = 6 extra per moved student.
- The moved group lies between the two old means: 4.5 below the first and 6 above the second. The old means differ by 4.5 + 6 = 10.5.
Extra practice: Count the middle child once
Five children stand in a line in height order. The leftmost three average 150 cm, the rightmost three average 147 cm, and all five average 148.5 cm. How tall is the middle child?
Hint
Draw A–B–C–D–E. The two groups are A–B–C and C–D–E: C is counted twice.
Worked solution — open after trying
The two totals sum to 450 + 441 = 891. This includes everyone once and the middle child once more. Subtract the whole total: 891 − 742.5 = 148.5 cm.
Mission 10
Not completed
Exit ticket
Complete these five questions without opening a hint. A perfect score unlocks your certificate.
Keep practicing
Use the feedback to decide which weight or total needs another look.
Certificate of Mastery
Weighted-Average Architect
This certifies that
can combine group totals and counts, recognize equal and unequal weights, use ratios as weights, and work backward to find a missing group size.
Teaching notes
Calculator group counts must be positive whole numbers.