Is it possible to find the combined standard deviation? Suppose I have 2 sets:
Set A: number of items $n= 10$, $\mu = 2.4$ , $\sigma = 0.8$
Set B: number of items $n= 5$, $\mu = 2$, $\sigma = 1.2$
I can find the combined mean ($\mu$) easily, but how am I supposed to find the combined standard deviation?
 A: So, if you just want to have two of these samples brought together into one you have:
$s_1 = \sqrt{\frac{1}{n_1}\Sigma_{i = 1}^{n_1} (x_i - \bar{y}_1)^2}$
$s_2 = \sqrt{\frac{1}{n_2}\Sigma_{i = 1}^{n_2} (y_i - \bar{y}_2)^2}$
where $\bar{y}_1$ and $\bar{y}_2$ are sample means and $s_1$ and $s_2$ are sample standard deviations.
To add them up you have:
$s = \sqrt{\frac{1}{n_1 + n_2}\Sigma_{i = 1}^{n_1 + n_2} (z_i - \bar{y})^2}$
which is not that straightforward since the new mean $\bar{y}$ is different from $\bar{y}_1$ and $\bar{y}_2$:
$\bar{y} = \frac{1}{n_1 + n_2}\Sigma_{i = 1}^{n_1 + n_2} z_i = \frac{n_1 \bar{y}_1 + n_2 \bar{y}_2}{n_1 + n_2}$
The final formula is:
$s = \sqrt{\frac{n_1 s_1^2 + n_2 s_2^2+ n_1(\bar{y}_1-\bar{y})^2 +n_2(\bar{y}_2-\bar{y})^2}{n_1 + n_2 }}$
For the commonly-used Bessel-corrected ("$n-1$-denominator") version of standard deviation, the results for the means are as before, but 
$s = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2 + n_1(\bar{y}_1-\bar{y})^2 +n_2(\bar{y}_2-\bar{y})^2}{n_1+n_2 - 1} }$
You can read more info here: http://en.wikipedia.org/wiki/Standard_deviation
A: This obviously extends to $K$ groups:
$$ s = \sqrt{ \frac{\sum_{k=1}^K (n_k-1)s_k^2 + n_k(\bar{y}_k-\bar{y})^2} {(\sum_{k=1}^K n_k) -1} }$$
