01 · Recover totals
Five observations have mean 3 and descriptive SD 2. Find T and U.
Hint
Use U = n(SD² + mean²).
Worked solution
T = 15; U = 5(4 + 9) = 65.
Understand · explore · practise
Correcting and combining summary statistics
Recover sums from means and standard deviations, correct errors, add or remove observations and combine disjoint datasets without averaging their standard deviations.
Before you startMean, variance, standard deviation and combined means.
01 / Keep a three-part record
Count, sum and squared sum have different jobs.
Write n, T = Σx and U = Σx².
Then mean = T/n and descriptive variance = U/n − (T/n)².
Work with these totals when data are corrected or combined. A mean alone loses the count; a standard deviation alone loses both the count and the location.
Correct both totals separatelyExplore
Start with values 2, 4, 6, 8: count 4, sum 20, squared sum 120. Replace the recorded value 8 by c.
Sum = 20 − 8 + 8 = 20.
Squared sum = 120 − 8² + 8² = 120.
Mean = 5; variance = 5; SD ≈ 2.236068.
A replacement keeps n = 4. Adding or removing a whole observation would also change the count.
02 / Recover the underlying totals
Square the SD to obtain the variance.
Six observations have mean 5 and descriptive SD 2.Worked example
T = n×mean = 6×5 = 30
Recover the ordinary sum.
Variance = 2² = 4
Do not insert SD directly into the variance formula.
U = n(variance + mean²) = 6(4 + 25) = 174
Recover the squared sum.
02 · Supplied variance
Ten observations have mean 4 and descriptive variance 9. Find U.
Hint
Nine is already a variance.
Worked solution
U = 10(9 + 16) = 250. Do not square nine again.
03 / Replace one recorded value
Subtract the old square and add the new square.
Tnew = Told − a + b; Unew = Uold − a² + b².
Here a was recorded and b is the corrected replacement. The count stays the same.
Watch: the sum and squared sum need different corrections
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Correct an error
Four recorded values have T = 30 and U = 380. One value was written as 18 instead of 8. Find the corrected mean and descriptive variance.
Hint
Replace 18 by 8 in both totals.
Worked solution
Tnew = 30 − 18 + 8 = 20. Unew = 380 − 324 + 64 = 120. Mean = 5; variance = 120/4 − 25 = 5.
04 · Correct another error
For n = 5, T = 25, U = 165, a recorded 9 should be 7. Find corrected mean and variance.
Hint
The squared-sum correction is −81 + 49.
Worked solution
Tnew = 23; Unew = 133. Mean = 23/5 = 4.6; variance = 133/5 − (23/5)² = 136/25 = 5.44.
04 / Adding and removing change n
Distinguish a replacement from a deletion.
05 · Remove an observation
For the data 2, 4, 6, 8, remove 8. Find the new n, T, U and variance.
Hint
Subtract one from the count as well.
Worked solution
n = 3; T = 12; U = 56. Mean = 4; variance = 56/3 − 16 = 8/3.
06 · Add an observation
Add 10 to the original data 2, 4, 6, 8. Find the new mean and variance.
Hint
n = 5; add 10 to T and 100 to U.
Worked solution
T = 30; U = 220. Mean = 6; variance = 220/5 − 36 = 8.
07 · Wrong square correction
Why is subtracting (18 − 8)² = 100 from U wrong when correcting 18 to 8?
Hint
The required change is a difference of squares, not a squared difference.
Worked solution
The correct reduction is 18² − 8² = 324 − 64 = 260. In general a² − b² is not (a − b)².
05 / Add totals for disjoint groups
Recover each squared sum before combining.
A: n = 3, mean 4, variance 8/3. B: n = 2, mean 10, variance 4. Variances use denominator n.Worked example
TA = 12; UA = 3(8/3 + 16) = 56
Recover group A’s sums.
TB = 20; UB = 2(4 + 100) = 208
Recover group B’s sums.
Combined n = 5, T = 32, U = 264
The groups are disjoint and measured in the same units.
Mean = 6.4; variance = 264/5 − (32/5)² = 296/25 = 11.84
The combined SD is √296/5 ≈ 3.441.
08 · Constant groups
A is 0, 0 and B is 10, 10. Each has SD zero. Find their combined SD.
Hint
The combined mean is five.
Worked solution
Squared deviations are all 25, so descriptive variance = 25 and SD = 5. Averaging the group SDs would incorrectly give zero.
09 · Overlap
Why must you not add the sums for a whole class and for the girls within that same class as if they were separate groups?
Hint
One group is contained in the other.
Worked solution
That would count the girls twice. Use disjoint groups or remove the overlap using enough information.
06 / Separate within-group and between-group variation
Group means can be far apart even when each group is tight.
Combined variance = Σ nᵢ[vᵢ + (mᵢ − m)²] / Σnᵢ
m is the combined mean; mᵢ and vᵢ are each group’s mean and descriptive variance.
The first term measures spread within groups. The second measures the separation of their means from the overall mean. This explains why simply averaging group variances usually misses part of the variation.
10 · Explain the constant example
For groups 0, 0 and 10, 10, which part of the formula supplies the combined variance?
Hint
Both within-group variances are zero.
Worked solution
All variation is between groups: their means are each five from the combined mean. Weighted squared distances therefore give variance 25.
11 · Equal group means
If disjoint groups have the same mean, does their combined descriptive variance equal the count-weighted average of their variances?
Hint
The between-group distances are then zero.
Worked solution
Yes, for these descriptive variances and compatible units. Their SDs still should not generally be averaged; take the square root after combining variances.
07 / Check the denominator convention
Recover Sxx differently when the supplied variance uses n − 1.
If supplied s² = Sxx/(n − 1), recover Sxx = (n − 1)s² and then U = Sxx + n mean². Do not multiply s² by n. Once you have n, T and U, combine them and use the denominator required by the final question.
12 · A sample-variance summary
A group has n = 3, mean 4 and sample variance s² = 4 using n − 1. Recover T and U.
Hint
Sxx = 2×4.
Worked solution
T = 12; Sxx = 8; U = 8 + 3×16 = 56.
13 · A final sample variance
The combined dataset in the worked example has n = 5, T = 32 and U = 264. Find its sample variance using n − 1.
Hint
Sxx = U − T²/n.
Worked solution
Sxx = 264 − 1024/5 = 296/5. Divide by four: sample variance = 74/5 = 14.8.
08 / Respect rounded summaries
Reconstructed totals may themselves be approximate.
If means or SDs were rounded, the recovered T and U are estimates. Small differences in rounded numbers can affect a variance calculation. Preserve precision where possible and check that the final variance is non-negative.
14 · Rounding and exact claims
A source reports a mean and SD to one decimal place. Can you claim the recovered squared sum is necessarily exact?
Hint
The rounded summaries do not uniquely determine the unrounded values.
Worked solution
No. Treat it as an estimate based on the reported summaries and state the precision limitation.
09 / Update the records before the summaries
Recompute after every correction.
Record n, T and U; make the appropriate changes to all three; then recompute the mean and requested variance or SD. Check disjoint groups, common units, denominator conventions and rounding.
Section 1 of 9 · Keep a three-part record