Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question

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 $v$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.