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Given a list of numbers $L$, we can compute for the mean of the numbers, the standard deviation and then the median. Then use the (non-parametric) skew to compute for the skewness using: $$S = \dfrac{\mu-v}{\sigma}$$ where $\mu$ is the mean of the numbers in $L$, $v$ is the median of the numbers in $L$ and $\sigma$ is the standard deviation of the numbers in $L$.

So say, we wish to reverse the process. Instead of being presented with a list and computing the skewness $S$, we are only given the sum $\mathcal{S}$ of the numbers in the list, $n$ as the number of numbers in the list, and its skewness $S$. Can we come up with a list of numbers $L$ (or a set as order doesn't matter) such that it has $n$ numbers, it has a sum equal to $\mathcal{S}$ and has a skewness value equal to $S$? Is this even possible? I can't seem to find a source anywhere about how to do this.

So for example if a list of $n=9$ numbers has a sum of $81$ and has skewness of $0.43$. All I know would be the mean to be $\mu=9$ but I'm left under the dark after that. Any suggestions or tips on how to proceed? Thanks so much!

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You have a large amount of freedom: for example the scale of the dispersion is unspecified.

You know the median is less than the mean, so let's arbitrarily choose the median to be $8$. Perhaps we can do this with only three numbers: $x_{(1)},x_{(2)},x_{(3)}$ with $x_{(2)}=8$ and mean $\frac{x_{(1)}+x_{(2)}+x_{(3)}}{3}=9$. We minimise $\sigma$ and maximise your skewness when $x_{(1)}=8$ and $x_{(3)}=11$ giving a population standard deviation of $\sqrt{2}$ and a skewness on your definition of about $0.7071$. That is more than $0.43$, so we can afford to increase the standard deviation and it seems that $x_{(1)}=6.78661, x_{(2)}=8, x_{(3)}=12.21339$ with a population standard deviation of about $2.32558$ gives your $0.43$ target figure.

You wanted nine rather than three numbers: that is not a problem as we can just repeat each value three times, though you might decide you want to spread out the nine values more, or to have a different difference between the mean and median. There were other arbitrary choices there, and you might want to adapt the process, for example perhaps to use a so-called sample standard deviation.

An algorithm which says find the distribution which maximises the magnitude of your skewness figure, and then relax the data to increase the standard deviation to hit the target you want, should work so long as there is a solution. Sticking to odd-number sample sizes $n$ so the median is simple, I think you can get a solution so long as the magnitude of your target skewness is no more than $\sqrt{1-\frac{2}{n+1}}$.

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