Class 11 Statistics Notes · GSEB

Combined Standard Deviation

Combined Standard Deviation — merge two groups into one spread using the within-plus-between variance decomposition. GSEB Class 11 Statistics notes with the four-step order, formulas and solved sums.

Last updated: 5 Oct 2026

Notes

Combining Two Groups

Words marked with a dotted underline carry the question they usually appear in — tap them to reveal it.

“Uncontrolled variation is the enemy of quality.”— Edward Deming

Combined standard deviation (s_c)

The standard deviation of the new group formed by pooling the observations of two groups — obtained from (n₁, x̄₁, s₁) and (n₂, x̄₂, s₂) without ever touching the raw data. Its partner is the combined mean, x̄c, the average of the pooled group.

Only one measure of dispersion can be pooled

Only the standard deviation supports an algebraic combination. There is no combined range, no combined quartile deviation and no combined mean deviation — pooled groups simply cannot be handled that way, which is also why the standard deviation is called the most suitable measure for further algebraic work.

The Two Formulas

One formula pools the averages, the other pools the spreads. Both need the same four inputs per group: n, x̄ and s.

Combined mean — step 1

\bar{x}_c = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}

Combined standard deviation — the last step

s_c = \sqrt{\frac{n_1(s_1^2 + d_1^2) + n_2(s_2^2 + d_2^2)}{n_1 + n_2}}

Deviation of group 1

d_1 = \bar{x}_1 - \bar{x}_c

Deviation of group 2

d_2 = \bar{x}_2 - \bar{x}_c

What each symbol stands for

n₁, n₂ = number of observations in the two groups · x̄₁, x̄₂ = their means · s₁, s₂ = their standard deviations · d₁, d₂ = how far each group mean sits from the combined mean.

Within-Group Spread + Between-Group Gap

s² — the within-group half

How far the observations sit from their own group mean. Two shifts can each be very steady internally and still sit far apart from each other.

d² — the between-group half

How far each group mean sits from the combined mean. Pooling two groups whose averages differ always pushes the combined spread up, even when both groups are tidy.

The decomposition in one line

Combined variance = within-group s² + between-group d², averaged with the group sizes as weights. When the two group means coincide, d = 0 and the formula collapses into a simple weighted average of the two variances — the gap between the means is the only thing the second term ever adds.

⭐ The Fixed Order of Steps

x̄c first, d₁ and d₂ second, s_c last

The order is not a suggestion: d₁ = x̄₁ − x̄c and d₂ = x̄₂ − x̄c cannot even be written down before x̄c exists. Computing the d values against a guessed or half-finished centre is the classic trap of this topic — the number that comes out looks perfectly reasonable and is wrong all the way to the answer.

The ladder below enforces exactly this order — steps refuse to open early
Order	Compute	Because
1	x̄c = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂)	It is the centre every later deviation is measured from
2	d₁ = x̄₁ − x̄c and d₂ = x̄₂ − x̄c	Both differences are defined only against x̄c
3	n₁(s₁² + d₁²) + n₂(s₂² + d₂²)	Needs s₁², s₂² (given) and d₁², d₂² (from step 2)
4	s_c = √[ that sum ÷ (n₁ + n₂) ]	One square root, at the very end

When Variances Are Given Instead of SDs

Ex. 4.6, Q2 — 16 and 25 are already s²

Sometimes the table prints the variance of each group (16 and 25, say) rather than its standard deviation. Those numbers go straight into the formula as s₁² and s₂² — do not take a square root. Keep working with the variances and take the one square root at the end, on the final combined figure.

Section A: n = 10, mean 25, variance 16 · Section B: n = 40, mean 20, variance 25

x̄c = (10 × 25 + 40 × 20)/50 = 1050/50 = 21 minutes

d₁ = 25 − 21 = +4 · d₂ = 20 − 21 = −1

s_c² = [10(16 + 16) + 40(25 + 1)]/50 = (320 + 1040)/50 = 27.2

s_c = √27.2 = 5.22 minutes — the only square root in the whole sum

Two Curves, One Pool

Illustration 22 in pictures: groups of five (1, 3, 5, 7, 9) and (2, 4, 6, 8, 10), both with variance 8 and means at 5 and 6. Merge them and watch which band appears.

Both groups have the same variance (8.00) and their means are only one unit apart — but until x̄c exists there is nothing to measure that gap against.

What the bands are really saying

The amber band is spread the groups already had; the rose band is spread they gain purely because their averages sit apart; the emerald band is the sum. Same-scale bars: here the between-group part is only 0.25 because the two means are a single unit apart — push the means further apart and it grows without any change in the groups themselves.

The Four-Step Ladder — Illustration 23

Two shifts of a factory: Shift I has 60 workers averaging 25 minutes (SD 5), Shift II has 40 workers averaging 20 minutes (SD 3). Click the steps in order — later ones stay locked.

Four steps, one fixed order — seed: n₁ 60, x̄₁ 25, s₁ 5 · n₂ 40, x̄₂ 20, s₂ 3

Step 1 is the only entry point: no step beyond the next one opens, so d₁ and d₂ can never be computed before x̄c exists.

Why the order is drilled

Step 1 gives the pooled centre of gravity for 100 workers. Only after it exists can a shift be said to sit “+2” or “−3” minutes away from that centre — and only then can those gaps be squared and folded into the variances.

First Order and Second Order Averages

Two tiers of averages

Mean, median and mode are the first order averages, whereas measures of dispersion are the second order averages — a dispersion measure is computed from an average, so it always sits one level above it.

First order

Mean · Median · Mode

Read straight off the observations — they locate the centre.

Second order

Range · QD · Mean Deviation · Standard Deviation

Built out of an average — they describe how far the observations sit from it.

Solved Example — Illustration 23

Solved Example

Problem

A factory makes items in two shifts — Shift I: 60 workers, mean manufacturing time 25 minutes, SD 5 minutes Shift II: 40 workers, mean manufacturing time 20 minutes, SD 3 minutes Find the combined standard deviation.

Solution

x̄c = 23 minutes · s_c = √24.6 = 4.9598 ≈ 4.96 minutes

A note on the printed working

The book's printed substitution for this illustration carries stray minus signs in two places — the formula itself is n₁(s₁² + d₁²) + n₂(s₂² + d₂²) over (n₁ + n₂), all plus signs, and the working above follows that.

Worked Example Gallery

Combined Standard Deviation — solved practicals

3 problems

Raw two groups — G₁ and G₂ (Illustration 22)

Full question

Five observations in each of two groups G₁ and G₂ are given below :

G₁ : 1, 3, 5, 7, 9

G₂ : 2, 4, 6, 8, 10

Find the mean and variance of both the groups. Hence obtain the combined standard deviation from it.

Both groups contribute 5 × 8.25 to the pooled sum of squares
Group	n	x̄	s²	d	s² + d²
G₁	5	5	8	−0.5	8.25
G₂	5	6	8	+0.5	8.25
Combined	10	5.5	—	—	8.25

x̄c = 5.5, s_c = √8.25 = 2.87

x̄₁ = 5, s₁² = 8 · x̄₂ = 6, s₂² = 8 · n₁ = n₂ = 5.
x̄c = (5 × 5 + 5 × 6)/10 = 55/10 = 5.5.
d₁ = 5 − 5.5 = −0.5 · d₂ = 6 − 5.5 = +0.5.
s_c = √[(5(8 + 0.25) + 5(8 + 0.25))/10] = √(82.5/10) = √8.25 = 2.8723 ≈ 2.87.

Pooling the ten raw values (1, 3, 5, 7, 9, 2, 4, 6, 8, 10) gives the same mean and SD — the book’s own activity check.

Given summaries — two shifts (Illustration 23)

Full question

A factory manufactures certain items in two shifts. The information regarding the time taken by workers to manufacture the items is given below. Using the following information, find the combined standard deviation :

	Shift I	Shift II
No. of workers	60	40
Mean manufacturing time (minutes)	25	20
Standard deviation (minutes)	5	3

Shift	n	x̄	s	d	n(s² + d²)
I	60	25	5	+2	60(25 + 4) = 1740
II	40	20	3	−3	40(9 + 9) = 720

x̄c = 23 min, s_c = √24.6 = 4.96 minutes

x̄c = (60 × 25 + 40 × 20)/100 = 23 min.
d₁ = +2 · d₂ = −3.
60(25 + 4) = 1740 · 40(9 + 9) = 720.
s_c = √(2460/100) = √24.6 = 4.96 min.

Variances given — two firms (Illustration 26, part 3)

Full question

The following results are obtained on the basis of daily wages (in ₹) paid to workers of two firms A and B :

	Firm A	Firm B
No. of workers	20	30
Mean daily wages (₹)	250	400
Standard deviation of wages (₹)	10	12

Answer the following questions using the above information.

3. Find the combined mean and combined standard deviation for the two firms.

Firm	n	x̄	s²	d	n(s² + d²)
A	20	₹250	100	−90	20 × 8200 = 164000
B	30	₹400	144	+60	30 × 3744 = 112320

x̄c = ₹340, s_c ≈ ₹74.34

The variances are already s₁² = 100 and s₂² = 144 — no square root here.
x̄c = (20 × 250 + 30 × 400)/50 = 17000/50 = ₹340.
d₁ = 250 − 340 = −90 · d₂ = 400 − 340 = +60.
20(100 + 8100) = 164000 · 30(144 + 3600) = 112320.
s_c = √(276320/50) = √5526.4 = 74.3398 ≈ ₹74.34.

Quick Reference

Idea	Statement
Combined mean	x̄c = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂) — always step 1
Combined SD	s_c = √[ (n₁(s₁² + d₁²) + n₂(s₂² + d₂²)) / (n₁ + n₂) ]
Order of work	d₁ and d₂ are computed only after x̄c is found
The two halves	Combined variance = within-group s² + between-group d²
Variances printed	Use s² as it stands (Ex. 4.6 Q2: 16 and 25) — no square root until the final step
Means coincide	d = 0, and the formula drops to a weighted average of the two variances
Only SD combines	No combined range, quartile deviation or mean deviation exists

Key Takeaways

Pooling two groups gives a combined mean and a combined SD — and among all dispersion measures only the standard deviation can be combined algebraically. So what: a pooled range or pooled quartile deviation simply does not exist; SD is the only tool that survives the merge.
x̄c = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂) and s_c = √[(n₁(s₁² + d₁²) + n₂(s₂² + d₂²))/(n₁ + n₂)], where d₁ = x̄₁ − x̄c and d₂ = x̄₂ − x̄c.
⭐ The order is fixed: x̄c first → d₁, d₂ → s² + d² → square root last. So what: measuring d before x̄c is the classic trap — the number looks plausible and poisons every step after it.
Combined variance has two halves: within-group s² and between-group d². If the group means coincide, d = 0 and the whole thing becomes a weighted average of the variances.
When variances are printed instead of SDs (Ex. 4.6 Q2: 16 and 25), take them as they are and square-root only the final answer.
Anchors: G₁/G₂ → x̄c 5.5, s_c 2.87 · two shifts → x̄c 23 min, s_c 4.96 min · two firms → x̄c ₹340, s_c ≈ ₹74.34. So what: one formula, three shapes of data — raw groups, printed summaries and printed variances — and the four steps never change order.