Variance

Question

Comments

E(x²)= $1^{2}\times (\frac{1}{2})+ 1^{2}\times (\frac{1}{2^{2}})+1^{2}\times (\frac{1}{2^{3}})$

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@akash

here x² can also take 2 values i.e 0 or 1, now,

probability to get x²=1, means probability of getting x=1, therfore

E(x²)= 1*P(x=1)

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@Anu sir,

yes exactly.