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Let $X$ be a discrete random variable with expected value $E(X)$. Further, suppose there is a $1/4$ probability of $X$ being exactly $2$ units away from $E(X)$, a $1/4$ probability of $X$ being exactly $3$ units away from $E(X)$, and a $1/2$ probability of $X$ being exactly $5$ units away from $E(X)$. What is $Var(X)$?

Natalie
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1 Answers1

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The variance is $E((X-\mu)^2)$, where $\mu=E(X)$ is the mean.

The required expectation is therefore $(1/4)(2^2)+(1/4)(3^2)+(1/2)(5^2)$.

André Nicolas
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  • There could be $6$ points. There are I think other combinations of weight distributions beside the symmetric one that would give the right mean. And the variance will not depend on the exact distribution, as long as the specifications are met. – André Nicolas Jun 20 '14 at 18:35
  • André, @Dan I read "there is a $1/4$ probability of $X$ being exactly $2$ units away from $E(X)$" as saying that there is a single mass of $\frac 14$ at distance $2$ from the mean, and didn't think of the possibility of a total mass of $\frac 14$ at distance $\pm 2$. I will delete my comment. – Dilip Sarwate Jun 20 '14 at 18:49