Discrete random variables — practice questions

(i) The sum of probabilities must be 1. $\sum P(X=x) = 1$ $p + 2p + 0.4 + 0.1 = 1$ $3p + 0.5 = 1$ $3p = 0.5$ $p = \frac{0.5}{3} = \frac{1}{6}$

So, the probabilities are $P(X=1) = \frac{1}{6}$ and $P(X=2) = 2 \times \frac{1}{6} = \frac{1}{3}$.

(ii) To find the expectation, $E(X)$, we use the formula $E(X) = \sum x P(X=x)$. $E(X) = (1 \times P(X=1)) + (2 \times P(X=2)) + (3 \times P(X=3)) + (4 \times P(X=4))$ $E(X) = (1 \times \frac{1}{6}) + (2 \times \frac{1}{3}) + (3 \times 0.4) + (4 \times 0.1)$ $E(X) = \frac{1}{6} + \frac{2}{3} + 1.2 + 0.4$ $E(X) = \frac{1}{6} + \frac{4}{6} + 1.6 = \frac{5}{6} + 1.6$ $E(X) = 0.8333... + 1.6 = 2.4333...$ As an exact fraction: $E(X) = \frac{5}{6} + \frac{16}{10} = \frac{5}{6} + \frac{8}{5} = \frac{25+48}{30} = \frac{73}{30}$. So, $E(X) = \frac{73}{30}$ (or 2.43 to 3 s.f.).

(i) First, calculate $E(Y)$. $E(Y) = \sum y P(Y=y)$ $E(Y) = (0 \times 0.1) + (1 \times 0.3) + (2 \times 0.4) + (3 \times 0.2)$ $E(Y) = 0 + 0.3 + 0.8 + 0.6 = 1.7$

Next, to find $Var(Y)$, we first need $E(Y^2)$. $E(Y^2) = \sum y^2 P(Y=y)$ $E(Y^2) = (0^2 \times 0.1) + (1^2 \times 0.3) + (2^2 \times 0.4) + (3^2 \times 0.2)$ $E(Y^2) = (0 \times 0.1) + (1 \times 0.3) + (4 \times 0.4) + (9 \times 0.2)$ $E(Y^2) = 0 + 0.3 + 1.6 + 1.8 = 3.7$

Now use the variance formula: $Var(Y) = E(Y^2) - [E(Y)]^2$. $Var(Y) = 3.7 - (1.7)^2$ $Var(Y) = 3.7 - 2.89 = 0.81$

(ii) We use the rules for linear transformations. Given $Z = 5Y - 2$, we have $a=5$ and $b=-2$.

For expectation: $E(Z) = E(5Y - 2) = 5E(Y) - 2$. $E(Z) = 5(1.7) - 2 = 8.5 - 2 = 6.5$

For variance: $Var(Z) = Var(5Y - 2) = 5^2 Var(Y)$. $Var(Z) = 25 \times Var(Y) = 25 \times 0.81$ $Var(Z) = 20.25$