| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Linear combinations of independent variables |
| Difficulty | Challenging +1.8 This is a Further Maths statistics question requiring knowledge that E(XY) = E(X)E(Y) for independent variables, and the variance formula Var(XY) = Var(X)Var(Y) + Var(X)[E(Y)]² + Var(Y)[E(X)]². While the calculations are straightforward once you know the formulas, this requires recall of non-standard results about products of independent random variables that go beyond typical A-level content, plus careful algebraic manipulation across multiple steps for 8 marks total. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02d Binomial: mean np and variance np(1-p)5.02m Poisson: mean = variance = lambda5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 3.6\) and \(E(Y) = 4\) | B1 | Both seen or implied in (a) or (b) |
| \(E(XY) = (3.6 \times 4) = 14.4\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(Var(X) = 2.52\) and \(Var(Y) = 4\) | B1 | Both si |
| \(E(X^2) = Var(X) + (E(X))^2 = 2.52 + 3.6^2 = 15.48\) | M1 A1 | Correct method for either \(E(X^2)\) or \(E(Y^2)\) |
| \(E(Y^2) = 4 + 4^2 = 20\) | A1 | |
| \(Var(XY) = E(X^2)E(Y^2) - (E(XY))^2 = 15.48 \times 20 - 14.4^2\) | m1 | Dep on previous M1. FT their 14.4, 15.48 and 20 for m1 only. cao |
| \(= 102.24\) | A1 | |
| [8] |
## 1(a)
$E(X) = 3.6$ and $E(Y) = 4$ | B1 | Both seen or implied in (a) or (b)
$E(XY) = (3.6 \times 4) = 14.4$ | B1 |
## 1(b)
$Var(X) = 2.52$ and $Var(Y) = 4$ | B1 | Both si
$E(X^2) = Var(X) + (E(X))^2 = 2.52 + 3.6^2 = 15.48$ | M1 A1 | Correct method for either $E(X^2)$ or $E(Y^2)$
$E(Y^2) = 4 + 4^2 = 20$ | A1 |
$Var(XY) = E(X^2)E(Y^2) - (E(XY))^2 = 15.48 \times 20 - 14.4^2$ | m1 | Dep on previous M1. FT their 14.4, 15.48 and 20 for m1 only. cao
$= 102.24$ | A1 |
| [8] |
---The random variable $X$ has the binomial distribution B(12, 0·3). The independent random variable $Y$ has the Poisson distribution Po(4). Find
\begin{enumerate}[label=(\alph*)]
\item $E(XY)$, [2]
\item Var$(XY)$. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 2018 Q1 [8]}}