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10 cards from this deck
μ\muμ or E(X)E(X)E(X)
E(X)=∑xx⋅p(x)E(X) = \sum_{x} x \cdot p(x)E(X)=∑xx⋅p(x)
Long-run average value over many trials
aE(X)+baE(X) + baE(X)+b
No, they are not equal in general
∑xg(x)⋅p(x)\sum_{x} g(x) \cdot p(x)∑xg(x)⋅p(x)
Var(X)\text{Var}(X)Var(X) or σ2\sigma^2σ2
E(X2)−[E(X)]2E(X^2) - [E(X)]^2E(X2)−[E(X)]2
Var(X)\sqrt{\text{Var}(X)}Var(X)
a2Var(X)a^2\text{Var}(X)a2Var(X)
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