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I'm given that n=2 and a simple table showing this...

x 0 1 5 p(x) .25 .25 .5

I found the sample distribution for the sample mean to be this... _ x 0 .5 1 2.5 3 5 _ p(x) .0625 .125 .0625 .25 .25 .25

I also discovered that the mean is 2.75, but I'm lost on how to find the sample variance for these points...I'd appreciate any feedback or tips!

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There are, unfortunately, two different quantities that are sometimes referred to as "sample variance". For $n$ values $y_1, \ldots, y_n$, the "biased sample variance" is $\dfrac{1}{n} \sum_{i=1}^n (y_i - \overline{y})^2$, and the "unbiased sample variance" is $\dfrac{1}{n-1} \sum_{i=1}^n (y_i - \overline{y})^2$, where in both cases $\overline{y} = \dfrac{1}{n} \sum_{i=1}^n y_i$ is the sample mean. You should check which one you are being asked about.

Since $n=2$ and there are just $3$ possible values, you just have $9$ outcomes to consider, each consisting of a possible value for $y_1$ and a possible value for $y_2$. Find the value of the sample variance for each.
For example, if $y_1 = 0$ and $y_2 = 5$, then $\overline{y} = (0+5)/2 = 2.5$ and the unbiased sample variance is $(0-2.5)^2 + (5-2.5)^2 = 12.5$.
Then the probability of the sample variance having value $v$ is the sum of the probabilities of the outcomes where the sample variance is $v$.

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