| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Functions of random variables |
| Difficulty | Challenging +1.2 This S4 question requires systematic enumeration of joint probabilities for functions of independent random variables, then computing covariance and conditional probability. While it involves multiple steps and careful bookkeeping across 9 cases, the techniques are standard for Further Maths statistics: independence allows straightforward probability multiplication, covariance uses the standard formula, and conditional probability is direct application of the definition. The conceptual demand is moderate—no novel insight required, just methodical application of S4 techniques. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(x\) | 0 | 1 | 2 |
| \(\mathrm { P } ( X = x )\) | 0.5 | 0.3 | 0.2 |
| \(y\) | 0 | 1 | 2 |
| \(\mathrm { P } ( Y = y )\) | 0.5 | 0.3 | 0.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{array}{c\ | cccc} WU & 0 & 1 & 2 & 4 \\ 0 & 0.25 & 0.09 & 0 & 0.04 \\ 1 & 0.3 & 0 & 0.12 & 0 \\ 2 & 0.2 & 0 & 0 & 0 \end{array}\) | B1 |
| M1 | Attempt to allocate each \((x,y)\) to correct cell | |
| A1 | At least 8 cells correct | |
| A1 | All 12 correct | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(U) = 0.09 + 0.24 + 0.16 = 0.49\) | M1 | Their totals |
| \(E(V) = 0.42 + 0.4 = 0.82\) | M1 | " " |
| \(E(UV) = 2 \times 0.12 = 0.24\) | M1 | Their table |
| \(\text{Cov}(U,V) = \text{"0.24"} - \text{"0.49"} \times \text{"0.82"} = -0.1618\) | M1,A1 | Allow \(-0.162\) |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(UV=0 \cap V=2)/P(V=2) = 0.2/0.2 = 1\) | M1A1 | Or verbal explanation. Either num/denom correct M1, but NOT e.g. \(\frac{0.88 \times 0.2}{0.2}\) |
| [2] |
# Question 2:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{array}{c\|cccc} WU & 0 & 1 & 2 & 4 \\ 0 & 0.25 & 0.09 & 0 & 0.04 \\ 1 & 0.3 & 0 & 0.12 & 0 \\ 2 & 0.2 & 0 & 0 & 0 \end{array}$ | B1 | For correct $U$ and $V$ values |
| | M1 | Attempt to allocate each $(x,y)$ to correct cell |
| | A1 | At least 8 cells correct |
| | A1 | All 12 correct |
| | **[4]** | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(U) = 0.09 + 0.24 + 0.16 = 0.49$ | M1 | Their totals |
| $E(V) = 0.42 + 0.4 = 0.82$ | M1 | " " |
| $E(UV) = 2 \times 0.12 = 0.24$ | M1 | Their table |
| $\text{Cov}(U,V) = \text{"0.24"} - \text{"0.49"} \times \text{"0.82"} = -0.1618$ | M1,A1 | Allow $-0.162$ |
| | **[5]** | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(UV=0 \cap V=2)/P(V=2) = 0.2/0.2 = 1$ | M1A1 | Or verbal explanation. Either num/denom correct M1, but NOT e.g. $\frac{0.88 \times 0.2}{0.2}$ |
| | **[2]** | |
---2 The independent discrete random variables $X$ and $Y$ can take the values 0,1 and 2 with probabilities as given in the tables.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
$x$ & 0 & 1 & 2 \\
\hline
$\mathrm { P } ( X = x )$ & 0.5 & 0.3 & 0.2 \\
\hline
\end{tabular}
\end{center}$\quad$\begin{tabular}{ | l | c | c | c | }
\hline
$y$ & 0 & 1 & 2 \\
\hline
$\mathrm { P } ( Y = y )$ & 0.5 & 0.3 & 0.2 \\
\hline
\end{tabular}
The random variables $U$ and $V$ are defined as follows:
$$U = X Y , V = | X - Y | .$$
(i) In the Printed Answer Book complete the table giving the joint distribution of $U$ and $V$.\\
(ii) Find $\operatorname { Cov } ( U , V )$.\\
(iii) Find $\mathrm { P } ( U V = 0 \mid V = 2 )$.
\hfill \mbox{\textit{OCR S4 2017 Q2 [11]}}