In an examination, Section A (50 students) scored a mean of 60 marks with SD 8, while Section B (50 students)

Worked Examples · Example 9

Q. In an examination, Section A (50 students) scored a mean of 60 marks with SD 8, while Section B (50 students) scored a mean of 55 marks with SD 6. Find the combined mean and combined standard deviation of both sections together.

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Step 1 — Find the combined mean.

\bar{x}_{12} = \dfrac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2} = \dfrac{50\times60+50\times55}{50+50} = \dfrac{3000+2750}{100} = \dfrac{5750}{100} = 57.5

Step 2 — Find $d_1$ and $d_2$ , the distance of each section's own mean from the combined mean.

d_1 = \bar{x}_1-\bar{x}_{12} = 60-57.5 = 2.5, \qquad d_2 = \bar{x}_2-\bar{x}_{12} = 55-57.5 = -2.5

Step 3 — Apply the combined-SD formula.

\sigma_{12}^2 = \dfrac{n_1(\sigma_1^2+d_1^2) + n_2(\sigma_2^2+d_2^2)}{n_1+n_2} = \dfrac{50(8^2+2.5^2) + 50(6^2+2.5^2)}{100} = \dfrac{50(64+6.25)+50(36+6.25)}{100}

= \dfrac{50\times70.25 + 50\times42.25}{100} = \dfrac{3512.5+2112.5}{100} = \dfrac{5625}{100} = 56.25

$\sigma_{12} = \sqrt{56.25} = 7.5$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CA Foundation 2026Set jan-20261 markMCQ
$Q. If two samples of size 30 and 20 have means as 55 and 60 and standard deviation 5 and 6 respectively then what would be the standard deviation of combined sample of size 50? (A) 5.95 (B) 5.90 (C) 5.85 (D) 5.80$
›Reveal solutionSolution

Combined SD $=\sqrt{\dfrac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}}$ where $d_i$ is each group mean's deviation from the combined mean; here it equals $\sqrt{35.4}\approx5.95$ .

Step 1 — Combined mean
$\bar x=\frac{30(55)+20(60)}{50}=\frac{1650+1200}{50}=\frac{2850}{50}=57.$
Step 2 — Deviations of each mean from the combined mean
$d_1=55-57=-2,\quad d_2=60-57=3.$
Step 3 — Pooled variance
$\sigma^2=\frac{30(5^2+(-2)^2)+20(6^2+3^2)}{50}=\frac{30(25+4)+20(36+9)}{50}=\frac{870+900}{50}=\frac{1770}{50}=35.4.$
Step 4 — Combined SD

$\sigma=\sqrt{35.4}\approx5.95.$ …
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CA Foundation 2023Set jun-20231 markMCQ
$Q. The mean and variance of a group of 100 observations are 8 and 9, respectively. Out of 100 observations, the mean and standard deviation of 60 observations are 10 and 2, respectively. Find the standard deviation of remaining 40 observations? (A) 4.5 (B) 3.5 (C) 2.5 (D) 1.5$
›Reveal solutionSolution

The remaining 40 observations have variance 1.5 (option D).

Step 1 — Find the missing group mean

Total sum = 100 × 8 = 800; first group sum = 60 × 10 = 600, so remaining sum = 200 and its mean = 200/40 = 5.

Step 2 — Apply the combined-variance formula

With grand mean 8: d₁ = 10 − 8 = 2, d₂ = 5 − 8 = −3.
$n\sigma^2 = n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)$ $100(9) = 60(4 + 4) + 40(\sigma_2^2 + 9)$
Step 3 — Solve for σ₂²

$900 = 480 + 40\sigma_2^2 + 360 \Rightarrow 40\sigma_2^2 = 60 \Rightarrow \sigma_2^2 = 1.5$ …
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Q.In an examination, Section A (50 students) scored a mean of 60 marks with SD 8, while Section B (50 students) scored a mean of 55 marks with SD 6. Find the combined mean and combined standard deviation of both sections together.

Concept understanding — Combined Standard Deviation of Groups

How it works

Method

Quick example

Step 1 — Combined mean

Step 2 — Deviations of each mean from the combined mean

Step 3 — Pooled variance

Step 4 — Combined SD

Step 1 — Find the missing group mean

Step 2 — Apply the combined-variance formula

Step 3 — Solve for σ₂²

Q.In an examination, Section A (50 students) scored a mean of 60 marks with SD 8, while Section B (50 students) scored a mean of 55 marks with SD 6. Find the combined mean and combined standard deviation of both sections together.

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