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10 questions from this quiz
Takes only specific, countable values
The value with the greatest probability
P(X≤M)≥12P(X \leq M) \geq \frac{1}{2}P(X≤M)≥21 and P(X≥M)≥12P(X \geq M) \geq \frac{1}{2}P(X≥M)≥21
E[X]=∑∀xxP(X=x)E[X] = \sum_{\forall x} xP(X = x)E[X]=∑∀xxP(X=x)
Var(X)=E[X2]−(E[X])2\text{Var}(X) = E[X^2] - (E[X])^2Var(X)=E[X2]−(E[X])2
No, they are different quantities
μ=n+12\mu = \frac{n+1}{2}μ=2n+1
It doesn't affect variance at all
E(X+Y)=E(X)+E(Y)E(X + Y) = E(X) + E(Y)E(X+Y)=E(X)+E(Y)
Only when XXX and YYY are independent
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