# Is it possible to compute the variance without computing the mean first?

I have a list of values of a random variable $x \in \mathbb R$. Is it possible to find the varience $\overline{(x - \overline x)^2}$ without computing the mean $\overline x$ first? That is to process the list only once.

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You can use that the variance is $\overline{x^2} - \overline {x}^2$, which takes only one pass (computing the mean and the mean of the squares simultaneously), but can be more prone to roundoff error if the variance is small compared with the mean.