= Arithmetic mean
{title2=$\bar x$}
{wiki}
For a nonempty finite list of <real numbers> $x_1,\ldots,x_n$, the <arithmetic mean> is
$$
\bar x=\frac1n\sum_{i=1}^n x_i.
$$
It is the <weighted mean> with equal weights, giving each observation equal influence. For $(2,4,9)$, the <sum> is $15$ and the <arithmetic mean> is $5$. Replacing every value by $5$ retains the same total. The residuals $-3,-1,4$ sum to zero, so the mean is also the balancing point of these equally weighted observations.
The <arithmetic mean> minimizes the <sum> of squared deviations by the <weighted mean minimizes squared deviations> identity with all weights one. A large input can consequently have a substantial effect: replacing $9$ by $90$ changes the mean to $32$, while the <median> remains $4$. The choice of summary depends on what aspect of the data matters.
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