I would like to estimate the standard deviation of a population. The data I have is the size and mean of each of $k$ independent uniform subsamples: $(n_1, \mu_1), (n_2, \mu_2), \ldots, (n_k, \mu_k)$. Here, the $i$'th subsample has $n_i$ elements, and those $n_i$ elements have a mean of $\mu_i$.  What is the best way to utilize all $2k$ values to produce the best estimate of the population standard deviation $\sigma$?
I have a set of measurements which is partitioned into M partitions. However, I only have the partition sizes $N_i$ and the means $\bar{x}_i$ from each partition. Because all measurements are assumed to be from the same distribution, I believe I can estimate the mean of the population, $\bar{y}$, and standard deviation of the mean, $\sigma_{mean}$: $$ N=\sum_{i=1}^M N_i $$ $$ \bar{y} = \frac{1}{N}\sum_{i=1}^MN_i\bar{x}_i $$  $$ \sigma_{mean}=\sqrt{\frac{1}{N}\sum_i N_i(\bar{x}_i-\bar{y})^2} $$  My questions:   Am I right in my assumptions, that the mean $\bar{y}$ can be computed as above?  How can I find the standard deviation for the population, given only the means?  I read that the standard deviation of the population and standard deviation of the mean is related with $$ \sigma_{mean}=\frac{\sigma}{\sqrt{n}} \mbox{[1]} $$ where $n$ is the number of samples used in the computation of $\bar{x}_i$. So is it actually as simple as just multiplying $\sigma_{mean}$ with $\sqrt{n}$ if $n$ for all means are the same?   If it's that simple, what do I do if each $\bar{x}_i$ is computed using a different number of samples?    [1]
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