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Question:Suppose $X$ is a uniformly distributed random variable with possible values $1,2, \ldots, 10$. Compute the expected value and variance of $X$.

I have started with making a column ($x$ on the left and $y:=P(X=x)$ on the right);

EXPECTED VALUE: $$\begin{matrix} X & Y &\ \\ 1 & \frac{1}{10} & (1\times.10) +\\ 2 & \frac{1}{10} & (2\times.10) +\\ \vdots &\vdots&\vdots\\ 10 & \frac{1}{10}& (10\times.10)=\\ \ & \ & =5.5\end{matrix}$$

VARIANCE: $$((.10)-5.5)^2 + ((.20)-5.5)^2 +\cdots+ ((1)-5.5)^2) = 24.59$$

Is this the correct way of handling uniformly distributed random variables?

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  • $\begingroup$ What would "Y" be? $\endgroup$ – user228113 Apr 26 '15 at 1:28
  • $\begingroup$ Also, you might check here to learn the proper way to typeset your questions. $\endgroup$ – user228113 Apr 26 '15 at 1:30
  • $\begingroup$ Y is the probability; therefore 1/10 since uniformly distributed $\endgroup$ – user234475 Apr 26 '15 at 1:32
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The method for the expected value is correct, the one for the variance is not.

If $m$ is the expected value of $X$, the variance of $X$ is defined to be the expected value of $(X-m)^2$. In your case $$(1-5.5)^2\cdot0.10+(2-5.5)^2\cdot 0.10 + \cdots +(10-5.5)^2\cdot0.10=\cdots$$ Anoter way to calculate the variance of $X$ is calculating the expected value of $X^2$, say $k:=1^2\cdot0.10+2^2\cdot0.10+\cdots+10^2\cdot0.10$, and considering $k-m^2$.

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  • $\begingroup$ GOT IT! Thank you! Now I have a variance of 8.25 $\endgroup$ – user234475 Apr 26 '15 at 2:06

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