Using the standard formula, I always arrive at 2/3, but shouldn't the answer be 1?
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2What formula are you using? – Jeremy Miles Nov 23 '15 at 01:09
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4Why exactly "should" the answer be 1? – gung - Reinstate Monica Nov 23 '15 at 01:34
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4https://en.wikipedia.org/wiki/Bessel%27s_correction – Martin Capodici Nov 23 '15 at 04:38
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1I think one can be misled by Wolfram Alpha's "variance". Variance should be computed using their "population variance" function. ;) – Ilyes Hamdi Jan 23 '21 at 11:28
3 Answers
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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.
gung - Reinstate Monica
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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. – Mark L. Stone Nov 23 '15 at 03:11
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9@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. – gung - Reinstate Monica Nov 23 '15 at 04:22
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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.
Antoni Parellada
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The correct answer is $2/3$.
$\mu = 0$ (average of x).
$Var(x) = \frac{(1-0)^2 + (0-0)^2 + (-1-0)^2}{3} = \frac{2}{3}$
Denis Zubo
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This is only true if the data is uniformly distributed on 3 values, or if you compute the empirical variance. – ArnoV Jan 23 '21 at 11:34
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@user715586 Your point has already been made and answered. See gung's comment on Mark L. Stone's comment. – Nick Cox Jan 23 '21 at 11:56