The mean and standard deviation of a set of $n_{1}$ observations are $\bar{x}_{1}$ and $s_{1},$ respectively,while the mean and standard deviation of another set of $n_{2}$ observations are $\bar{x}_{2}$ and $s_{2},$ respectively. Show that the standard deviation of the combined set of $(n_{1}+n_{2})$ observations is given by $SD = \sqrt{\frac{n_{1}(s_{1})^{2}+n_{2}(s_{2})^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}(\bar{x}_{1}-\bar{x}_{2})^{2}}{(n_{1}+n_{2})^{2}}}$.

(N/A) Let the two sets of observations be $x_{i}$ $(i=1, 2, \ldots, n_{1})$ and $y_{j}$ $(j=1, 2, \ldots, n_{2})$.
The combined mean $\bar{x}$ is given by $\bar{x} = \frac{n_{1}\bar{x}_{1} + n_{2}\bar{x}_{2}}{n_{1} + n_{2}}$.
The variance of the combined set is $\sigma^{2} = \frac{1}{n_{1}+n_{2}} [\sum (x_{i}-\bar{x})^{2} + \sum (y_{j}-\bar{x})^{2}]$.
Using the identity $\sum (x_{i}-\bar{x})^{2} = \sum (x_{i}-\bar{x}_{1} + \bar{x}_{1}-\bar{x})^{2} = \sum (x_{i}-\bar{x}_{1})^{2} + n_{1}(\bar{x}_{1}-\bar{x})^{2} = n_{1}s_{1}^{2} + n_{1}d_{1}^{2}$,where $d_{1} = \bar{x}_{1}-\bar{x}$.
Similarly,$\sum (y_{j}-\bar{x})^{2} = n_{2}s_{2}^{2} + n_{2}d_{2}^{2}$,where $d_{2} = \bar{x}_{2}-\bar{x}$.
Substituting $d_{1} = \frac{n_{2}(\bar{x}_{1}-\bar{x}_{2})}{n_{1}+n_{2}}$ and $d_{2} = \frac{n_{1}(\bar{x}_{2}-\bar{x}_{1})}{n_{1}+n_{2}}$,we get:
$\sigma^{2} = \frac{n_{1}s_{1}^{2} + n_{2}s_{2}^{2}}{n_{1}+n_{2}} + \frac{n_{1}d_{1}^{2} + n_{2}d_{2}^{2}}{n_{1}+n_{2}}$.
Simplifying the term $\frac{n_{1}d_{1}^{2} + n_{2}d_{2}^{2}}{n_{1}+n_{2}}$ leads to $\frac{n_{1}n_{2}(\bar{x}_{1}-\bar{x}_{2})^{2}}{(n_{1}+n_{2})^{2}}$.
Thus,$SD = \sqrt{\frac{n_{1}s_{1}^{2}+n_{2}s_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}(\bar{x}_{1}-\bar{x}_{2})^{2}}{(n_{1}+n_{2})^{2}}}$.