Discrete Distributions and Expectation (AQA A-Level Further Maths): Quizzes

Takes only specific, countable values

The value with the greatest probability

$P(X \leq M) \geq \frac{1}{2}$ and $P(X \geq M) \geq \frac{1}{2}$

$E[X] = \sum_{\forall x} xP(X = x)$

$\text{Var}(X) = E[X^2] - (E[X])^2$

No, they are different quantities

$\mu = \frac{n+1}{2}$

It doesn't affect variance at all

$E(X + Y) = E(X) + E(Y)$

Only when $X$ and $Y$ are independent