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Suppose the set $S$ contains $n$ positive integers. If the mean $\mu$ of the elements is known, is there a method to finding the maximum possible value of the standard deviation $\sigma$ for $S$?

I know there have to be a finite number of subsets of the positive integers that have $n$ elements and mean $\mu$; each element must be less than or equal to the sum of every number in the set, $n\mu$, so we have at most $(n\mu)^n$ possibilities for $S$. However, I can't think of a method that can reliably maximize the spread of $S$. For context, I'm specifically working with a set of 1000 positive integers with $\mu=10$, so brute forcing it is not an option. Any help is appreciated.

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The most extreme case will be when $n-1$ integers take the minimum value $1$ and the remaining integer $n\mu-n+1$. Any other case can have the standard deviation increased by moving a pair of values further apart

This has mean $\mu$ and standard deviation (using the population $\frac1n$ method) of $(\mu -1)\sqrt{n-1}$

(If you insist on using the sample $\frac{1}{n-1}$ method then the standard deviation would instead be $(\mu -1)\sqrt{n}$)

In your example with $n=1000$ and $\mu=10$, this would be $999$ values of $1$ and $1$ value of $9001$, with a standard deviation of about $284.423$ (or $284.605$)

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