Standard Deviation for Combined Data

When two data sets (groups) are pooled into one combined data set, the mean and standard deviation of the combined set can be computed directly from the size, mean and standard deviation of each group, without going back to the individual observations. If a first group has $n_1$ items, mean $\bar{x}_1$ and standard deviation $\sigma_1$, and a second group has $n_2$ items, mean $\bar{x}_2$ and standard deviation $\sigma_2$, the mean of the combined data of size $n_1+n_2$ is the size-weighted average: $\bar{x}_c = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1+n_2}$ The standard deviation of the combined series is $\sigma_c = \sqrt{\frac{n_1\big(\sigma_1^2+d_1^2\big) + n_2\big(\sigma_2^2+d_2^2\big)}{n_1+n_2}}$ where $d_1 = \bar{x}_1 - \bar{x}_c$ and $d_2 = \bar{x}_2 - \bar{x}_c$ measure how far each group's own mean is offset from the combined mean. These $d^2$ correction terms are essential: simply averaging the two variances would understate the true spread of the pooled data whenever the two groups are centred at different means, since part of the combined spread then comes from the groups' means being apart, not only from the spread within each group.

Worked Example (Ex.1): Two samples of sizes $n_1=10$ and $n_2=20$ have means $\bar{x}_1=24,\bar{x}_2=45$ and standard deviations $\sigma_1=6,\sigma_2=11$. Combined mean: $\bar{x}_c = \frac{10(24)+20(45)}{30} = \frac{1140}{30} = 38$. Then $d_1 = 24-38=-14$ and $d_2=45-38=-7$. Combined variance $= \frac{10(6^2+14^2)+20(11^2+7^2)}{30} = \frac{10(232)+20(170)}{30} = \frac{2320+3400}{30} = \frac{5720}{30} \approx 190.67$, so the combined S.D. is $\sigma_c = \sqrt{190.67} \approx 13.81$. …