What is the combined mean of the two datasets?

Combined mean = (4×1 + 6×2)/(4+6) = (4+12)/10 = 16/10 = 1.6

What is the formula for combined variance?

σ²_combined = [n₁(σ₁²+d₁²) + n₂(σ₂²+d₂²)]/(n₁+n₂) where d_i = x̄_i − x̄_combined

What are d₁ and d₂?

d₁ = 1−1.6 = −0.6, d₁² = 0.36. d₂ = 2−1.6 = 0.4, d₂² = 0.16

What is the combined variance?

[4(13+0.36) + 6(1+0.16)]/10 = [4×13.36 + 6×1.16]/10 = [53.44+6.96]/10 = 60.40/10 = 6.04

Set of 4 observations mean=1 variance=13, set of 6 observations mean=2 variance=1 — combined variance of all 10

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MathStatistics

Q MCQ Statistics

A set of four observations has mean $1$ and variance $13$. Another set of six observations has mean $2$ and variance $1$. Then, the variance of all these $10$ observations is equal to:

A) $5.96$ B) $6.14$ C) $6.04$ D) $6.24$

✅ Correct Answer

C) 6.04

✓

Solution

Find the combined mean

$$\bar{x}=\frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2}=\frac{4\times1+6\times2}{4+6}=\frac{4+12}{10}=\frac{16}{10}=1.6$$

Compute the deviations d₁ and d₂

$d_1=\bar{x}_1-\bar{x}=1-1.6=-0.6\quad\Rightarrow\quad d_1^2=0.36$

$d_2=\bar{x}_2-\bar{x}=2-1.6=0.4\quad\Rightarrow\quad d_2^2=0.16$

Apply the combined variance formula

$$\sigma^2=\frac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}$$
$$=\frac{4(13+0.36)+6(1+0.16)}{10}$$
$$=\frac{4\times13.36+6\times1.16}{10}$$
$$=\frac{53.44+6.96}{10}=\frac{60.40}{10}=\boxed{6.04}$$

📘 Key Formula — Combined Variance

For two groups with sizes $n_1,n_2$, means $\bar{x}_1,\bar{x}_2$, variances $\sigma_1^2,\sigma_2^2$:

Combined mean: $\bar{x}=\dfrac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2}$

Combined variance: $\sigma^2=\dfrac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}$

where $d_i=\bar{x}_i-\bar{x}$ (deviation of group mean from combined mean).

💡

Important Concepts

Why Add d² to Group Variance? Each group's variance is measured about its own mean. When combining, we shift the reference to the overall mean. The extra $d_i^2$ term accounts for the displacement of each group's mean from the combined mean.

Special Case: Same Means If $\bar{x}_1=\bar{x}_2$, then $d_1=d_2=0$ and combined variance $=\dfrac{n_1\sigma_1^2+n_2\sigma_2^2}{n_1+n_2}$, a simple weighted average of the two variances.

Alternative Form The combined variance formula is equivalent to: $\sigma^2=\dfrac{\sum x_i^2}{N}-\bar{x}^2$ where $\sum x_i^2=n_1(\sigma_1^2+\bar{x}_1^2)+n_2(\sigma_2^2+\bar{x}_2^2)$ and $N=n_1+n_2$.

Exam Strategy Always compute combined mean first, then find $d_1,d_2$. Then substitute into the formula. Keep decimals careful — the four answer options (5.96, 6.04, 6.14, 6.24) are all close to each other, so precision matters.

FAQs

Why can't we just average the two variances directly?

⌄

Because the two groups have different means. Simple averaging ignores the spread introduced by the difference in group means. The $d_i^2$ term captures this additional source of variance in the combined dataset.

What is the alternative formula using Σx² and Σx²?

⌄

$\sum x_i^2 = n_1(\sigma_1^2+\bar{x}_1^2)+n_2(\sigma_2^2+\bar{x}_2^2) = 4(13+1)+6(1+4)=56+30=86$. Combined variance $=86/10 - (1.6)^2 = 8.6-2.56=6.04$ ✓. Both methods agree.

Can this formula extend to 3 or more groups?

⌄

Yes: $\sigma^2=\dfrac{\sum_i n_i(\sigma_i^2+d_i^2)}{\sum_i n_i}$ where $d_i=\bar{x}_i-\bar{x}_{\text{combined}}$.

What if variance of one group is 0?

⌄

If $\sigma_1^2=0$, all 4 observations are equal to their mean (= 1). Combined variance $= [4(0+0.36)+6(1+0.16)]/10 = [1.44+6.96]/10=0.84$. The formula still works correctly.

Is this from JEE Main 2026?

⌄

Yes, this question appeared in JEE Main 2026.

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