01 · Two groups
Six items have mean mass 14 g; nine other items have mean mass 24 g. Find the combined mean mass.
Hint
Add the two mass totals and the two counts.
Worked solution
(6×14 + 9×24)/15 = 300/15 = 20 g.
Understand · explore · practise
Combining means
Recover totals from group means, combine unequal groups correctly and update a mean when observations are added or removed. Original manual model and worked practice.
Before you startMean and weighted frequency calculations.
01 / Recover totals first
A group mean needs its group size.
Combined mean = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂)
This assumes disjoint groups measuring the same variable in compatible units.
Multiplying a mean by its observation count recovers the group total. Add totals and counts, then divide once.
Group size changes the weightExplore
Group A has mean 10 minutes. Group B has mean 20 minutes. These are invented groups of comparable observations.
Total = 5×10 + 5×20 = 150 minutes.
Combined mean = 150/10 = 15 minutes.
The groups are equally large, so the combined mean is halfway between their means.
The unweighted average of the two means is always 15, but it is not the overall per-observation mean when these group sizes differ.
02 / Combine unequal groups
Every observation receives the same weight.
Eight journeys average 12 minutes; twelve others average 17 minutes.Worked example
First total = 8×12 = 96 minutes
Recover the first group total.
Second total = 12×17 = 204 minutes
The larger group contributes more observations.
Combined mean = (96 + 204)/(8 + 12) = 15 minutes
The result lies closer to 17 than to 12 because that group is larger.
02 · Avoid averaging averages
A learner instead calculates (14 + 24)/2 = 19. Explain the error.
Hint
The two group sizes differ.
Worked solution
This gives each group equal weight rather than each item equal weight. The group of nine items should carry more weight.
03 / When does the shortcut work?
Equal sizes give equal weights.
Watch: recover totals before combining means
Pause, replay or seek freely. The notes explain the same idea and stay in view.
If the groups have equal positive sizes, their overall mean is the arithmetic average of their means. The shortcut also happens to agree if both means are identical. Otherwise unequal sizes require weighting.
03 · Equal sizes
Two disjoint groups of seven have means 8 and 12. Find the combined mean.
Hint
Both group weights are seven.
Worked solution
(7×8 + 7×12)/14 = 10.
04 · A reasonableness check
Both groups have positive sizes and means 8 and 12. Can the combined mean be 13?
Hint
A positive weighted average is between the two means.
Worked solution
No. It must lie between 8 and 12 (strictly between when the means differ and both counts are positive).
04 / More than two groups
Add all totals, not an average of averages.
The same method extends to any number of disjoint groups. Check for overlaps: if one group is included inside another, adding their counts double-counts observations.
05 · Three groups
Groups of 4, 6 and 10 observations have respective means 5, 10 and 14. Find their overall mean.
Hint
The total count is twenty.
Worked solution
Total = 4×5 + 6×10 + 10×14 = 220; mean = 220/20 = 11.
06 · Overlapping groups
A school reports the mean for all Year 12 pupils and separately the mean for its Year 12 maths pupils. Can you combine these as disjoint groups?
Hint
Maths pupils already appear in the year-group total.
Worked solution
No. Adding totals and counts would count those pupils twice. You need non-overlapping groups or enough information to remove the overlap.
05 / Find a missing group result
Use the combined total as an equation.
07 · Unknown mean
Ten observations have mean 7. After six others are included, the overall mean is 10. Find the second group’s mean.
Hint
Combined total = 16×10; subtract the first total.
Worked solution
Second total = 160 − 70 = 90. Second mean = 90/6 = 15.
08 · Unknown count
A group of eight has mean 10. A second group has mean 16. Their combined mean is 14. Find the second group size n.
Hint
Set (80 + 16n)/(8 + n) = 14.
Worked solution
80 + 16n = 112 + 14n, so 2n = 32 and n = 16.
06 / Add or remove observations
Change both the total and the count.
Five values have mean 12; a new value is 18.Worked example
Old total = 5×12 = 60
The mean alone is not a total.
New total = 60 + 18 = 78; new count = 6
The observation is additional.
New mean = 78/6 = 13
A value above the old mean raises it.
09 · Remove one value
Eight values have mean 15. A value of 22 is removed. Find the remaining mean.
Hint
Subtract 22 from 120, then divide by seven.
Worked solution
(8×15 − 22)/7 = 98/7 = 14.
10 · Direction without calculation
A new observation equal to the current mean is added. What happens to the mean?
Hint
The total rises by exactly one equal share.
Worked solution
It stays the same: if the old total is nm and the new value is m, the new mean is (nm + m)/(n + 1) = m.
07 / Check units and rounded summaries
A precise-looking answer can still be an estimate.
Convert units before adding totals. If the supplied means were rounded, the recovered totals and combined mean are approximate. Do not claim extra accuracy simply because the calculator shows many digits.
11 · Different units
Two durations average 90 seconds; three others average 2 minutes. Find the overall mean in seconds.
Hint
Convert two minutes to 120 seconds first.
Worked solution
(2×90 + 3×120)/5 = 540/5 = 108 seconds.
12 · Rounded means
Two group means are supplied to one decimal place. Is their weighted combination necessarily the exact mean of the original observations?
Hint
The original totals were not recovered exactly.
Worked solution
No. It is an estimate based on rounded summaries. State an appropriate precision and the limitation.
08 / Total divided by count
Keep the weighting visible.
- Check that groups are disjoint and measure comparable quantities.
- Convert units if needed.
- Recover each total using count × mean.
- Add totals and counts separately.
- Divide and interpret the result.
Section 1 of 8 · Recover totals first