Mathematics 9709 · AS & A Level · Linear combinations of random variables

Linear combinations of random variables — practice question

A random variable $X$ is given by the probability distribution in the table below, with $p$ as a constant. The possible values are $x = -1, 0, 1, 2, 4$ and the corresponding probabilities are $P(X = -1) = p$, $P(X = 0) = p$, $P(X = 1) = 2p$, $P(X = 2) = 2p$, and $P(X = 4) = 0.1$.
(i)[1]

Find $p$.

(ii)[2]

With $\text{E}(X) = 1.15$ given, determine $\operatorname{Var}(X)$.

Worked solution & mark scheme

This 3-mark question has a full step-by-step worked solution and mark scheme. One marking point: Solving $6p+0.1=1$ gives $p=0.15$

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