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How do I create a population (set of positive integers) given the size, mean and standard deviation? Like, if N = 7, mean is 50 and the standard deviation is 20, can I generate a population that fits that criteria?

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If $N = 7$ and the mean $\mu = \frac{\sum x}{N} = 50$, then $\sum x = 350$.

Recall that the variance of a population (not a sample) is defined as

$$\sigma^2 = \frac{\sum(x - \mu)^2}{N} = \frac{\sum x^2}{N} - \mu^2$$

Plugging in $N = 7$, $\mu = 50$, and $\sigma = 20$ gives:

$$20^2 = \frac{\sum x^2}{7} - 50^2$$

From which we get $\sum{x^2} = 20300$.

So now the problem becomes: Find 7 numbers that sum to 350 whose squares sum to 20300.

Through computer-assisted trial-and-error, I found that $[9, 42, 52, 54, 55, 57, 81]$ is one possible all-integer solution.

Dan
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    Subtracting your example from $100$ would give $91, 58, 48, 46, 45, 43, 19$ as another example, but there will be many more – Henry Jul 14 '22 at 09:27