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Statistics
OpenStudy (anonymous):

Find the Variance of a Discrete Random Variable For the # of H when we flip a coin twice in a discrete distribution: Data (respectively) X: 0,1,2 P(x): (1/4) (1/2) (1/4) mu=1 Please help me. Thank you!

OpenStudy (kirbykirby):

\(Var(X)=E(X^2)-[E(X)]^2=E(X^2)-\mu^2\) Now, \(E(X)=\mu \) And they already give you that value, which is 1 So, \(E(X^2)=\sum\limits_x x^2\cdot p(x)=0^2\left(\frac{1}{4} \right)+1^2\left(\frac{1}{2} \right)+2^2\left(\frac{1}{4} \right)\)

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