On this page
Section A scored a mean of 62 with a standard deviation of 8. Section B scored 72 with a standard deviation of 6. What is the standard deviation of the whole class? Many students answer with some average of 8 and 6. That answer is always too small when the means differ. This post derives the right formula, works three examples with checks, and ends with a routine.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "calculation of standard deviation, variance and mean deviation for grouped and ungrouped data". The CBSE 2026-27 curriculum puts variance and standard deviation in Class 11. NCERT's Class 11 chapter on statistics (2026-27 reprint) does not print a combined-variance formula, but its Example 16 uses the tool you need: σ² = (1/n)Σx² − x̄². Combined-group questions are built from that one line. Check your own year's syllabus.
Where does the combined formula come from?
Work with sums, not with the data.
For each group, Σx = n × mean.
From σ² = Σx²/n − mean², each group's Σx² = n(σ² + mean²).
Add the two Σx and the two Σx². Divide by n1 + n2.
Combined variance = (total Σx²)/(n1 + n2) − (combined mean)².
Tidying this up gives the textbook form:
σ² = [n1(σ1² + d1²) + n2(σ2² + d2²)] / (n1 + n2), where d1 and d2 are the distances of each group's mean from the combined mean.
An equivalent form shows why averaging fails:
σ² = (n1σ1² + n2σ2²)/(n1 + n2) + n1n2(x̄1 − x̄2)²/(n1 + n2)².
The first part is the weighted average of the variances, the spread inside the groups. The second part is the spread between the groups, and it is zero only when the two means are equal.
Worked example 1: two sections (sample numbers)
Section A: 30 students, mean 62, SD 8. Section B: 20 students, mean 72, SD 6. These are made-up classroom figures.
Combined mean = (30 × 62 + 20 × 72)/50 = (1860 + 1440)/50 = 66.
d1 = 62 − 66 = −4 and d2 = 72 − 66 = 6.
σ² = [30(64 + 16) + 20(36 + 36)]/50 = (2400 + 1440)/50 = 76.8.
σ = √76.8 ≈ 8.76.
Check with the second form: (30 × 64 + 20 × 36)/50 = 52.8, and 30 × 20 × 10²/50² = 24. 52.8 + 24 = 76.8. It agrees.
The weighted average of the SDs, (30 × 8 + 20 × 6)/50 = 7.2, is well short. The combined SD is larger than either section's because the two means are 10 marks apart.
Worked example 2: working backwards
Group 1 has 10 values with mean 20 and variance 9. Combined with group 2, there are 25 values with mean 23 and variance 19.2. Find the mean and SD of group 2.
n2 = 25 − 10 = 15.
Total Σx = 25 × 23 = 575. Group 1's Σx = 200, so group 2's mean = (575 − 200)/15 = 25.
Total Σx² = 25(19.2 + 529) = 25 × 548.2 = 13,705. Group 1's Σx² = 10(9 + 400) = 4090.
Group 2's Σx² = 13,705 − 4090 = 9615, so its variance = 9615/15 − 25² = 641 − 625 = 16, and its SD is 4.
This reverse type is where the sums method pays off: the formula with d1 and d2 is awkward to run backwards, but sums just subtract.
Worked example 3: equal means
Two groups have the same mean. One has 40 values with variance 25, the other 60 values with variance 16. The between-group term is zero, so the combined variance is the weighted average: (40 × 25 + 60 × 16)/100 = (1000 + 960)/100 = 19.6. This is the only case where a weighted average gives the right answer, and it averages variances, not SDs.
Which mistakes cost the most marks?
Averaging standard deviations. Combine variances, through the sums.
A simple average when the group sizes differ. Weight by n.
Dropping the d² terms, which only works when the means are equal.
Using σ instead of σ² when rebuilding Σx². Square first.
Dividing by n − 1. NCERT and JEE divide by n unless told otherwise.
For correcting a wrongly copied value with the same sums method, see our post on corrected mean and variance problems. For the basics, read mean, variance and standard deviation.
A practice routine
Dunlosky and colleagues (2013) reviewed ten study techniques and rated practice testing and distributed practice as high utility, and rereading as low. So spread this over a week and test yourself.
Day 1: derive the combined formula from Σx² = n(σ² + mean²) on blank paper.
Day 3: three forward problems; check each with both forms.
Day 5: two reverse problems like example 2.
Day 7: before calculating, predict whether the combined SD will be above both group SDs, then check.
If the idea of spread about the mean still feels slippery, the Fermi Concept Library covers every concept in mathematics, each with its own lesson, so you can revisit variance before the formulas.
Today, change Section B's mean in example 1 to 62 and predict the new combined SD before you work it out.
