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

Question

Accepted Answer

> [!TLDR]
> 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

$$\sigma=\sqrt{35.4}\approx5.95.$$ …

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 — Combined mean

Step 2 — Deviations of each mean from the combined mean

Step 3 — Pooled variance

Step 4 — Combined SD

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