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Using the standard formula, I always arrive at 2/3, but shouldn't the answer be 1?

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    $\begingroup$ What formula are you using? $\endgroup$ Commented Nov 23, 2015 at 1:09
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    $\begingroup$ Why exactly "should" the answer be 1? $\endgroup$ Commented Nov 23, 2015 at 1:34
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    $\begingroup$ en.wikipedia.org/wiki/Bessel%27s_correction $\endgroup$ Commented Nov 23, 2015 at 4:38
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    $\begingroup$ I think one can be misled by Wolfram Alpha's "variance". Variance should be computed using their "population variance" function. ;) $\endgroup$ Commented Jan 23, 2021 at 11:28

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You can think of the variance as the average squared deviation from the mean. Your mean is clearly $0$. With three data, you have three deviations (and thus three squared deviations):
\begin{align} &\ (1-0)^2 &=& &1^2& &=& &1 \\ &(-1-0)^2 &=& &-1^2& &=& &1 \\ &\ (0-0)^2 &=& &0^2& &=& &0 \end{align} The mean of $\{1, 1, 0\}$ is clearly $^2/_3$.

I think your intuition is neglecting the fact that there is a $0$ deviation in the set.

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  • $\begingroup$ Your answer assumes that each of 1, -1, 0 are equally likely. If 1 and -1 each have probability 1/2 of occurring, then the variance is 1. $\endgroup$ Commented Nov 23, 2015 at 3:11
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    $\begingroup$ @MarkL.Stone, this is just the variance of a set of numbers. It isn't even an estimate of the variance of the population from which the numbers were drawn. We can get fancier, but my interpretation is that that isn't what the question is about. $\endgroup$ Commented Nov 23, 2015 at 4:22
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You are thinking of range intuitively, and equating it with variance. But if you think of the mathematical formulation of the concept, you'll see that what leads you to some mistaken conclusion is that the zero point in your data is identical to the mean (zero), and consequently, one of the squared distances is zero, bringing down the calculation of the variance.

In other words, your data is not so spread on either side of the mean as to be $1$, because one of the data points lies precisely on the mean.

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The correct answer is $2/3$.

$\mu = 0$ (average of x).

$Var(x) = \frac{(1-0)^2 + (0-0)^2 + (-1-0)^2}{3} = \frac{2}{3}$

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  • $\begingroup$ This is only true if the data is uniformly distributed on 3 values, or if you compute the empirical variance. $\endgroup$ Commented Jan 23, 2021 at 11:34
  • $\begingroup$ @user715586 Your point has already been made and answered. See gung's comment on Mark L. Stone's comment. $\endgroup$
    – Nick Cox
    Commented Jan 23, 2021 at 11:56

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