| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Derived random variables from samples |
| Difficulty | Challenging +1.2 This S4 question requires constructing a joint probability distribution table from derived random variables, then applying standard formulas for covariance, independence testing, and conditional probability. While it involves multiple steps and careful enumeration of cases, the techniques are all routine for Further Maths statistics—no novel insight or complex problem-solving is required, making it moderately above average difficulty. |
| Spec | 2.03a Mutually exclusive and independent events5.03a Continuous random variables: pdf and cdf5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X \backslash Y\) table with values: \((0,2): \frac{1}{9}\); \((0,4): \frac{1}{9}\); \((1,3): \frac{2}{9}\); \((1,5): \frac{2}{9}\); \((2,4): \frac{2}{9}\) (remaining cells 0) | M1, A2, [3] | Consider at least 6 \((U,V)\) pairs; all correct. A1 for at least 9 correct cells |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \frac{8}{9}\) or \(E(Y) = 4\) | B1 | |
| \(E(XY) = \frac{32}{9}\) | B1 | CWO |
| \(\text{Cov}(X,Y) = \frac{32}{9} - 4 \times \frac{8}{9} = 0\) | M1A1, [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Consider e.g. \(P(X=1) \times P(Y=2) = \frac{4}{9} \times \frac{1}{9}\) | M1 | SC B1 no, from incorrect covariance |
| \(\neq P(X=1, Y=2) = 0\) so no | A1, [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{2}{9} \div \frac{4}{9} = \frac{1}{2}\) | M1A1, [2] |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X \backslash Y$ table with values: $(0,2): \frac{1}{9}$; $(0,4): \frac{1}{9}$; $(1,3): \frac{2}{9}$; $(1,5): \frac{2}{9}$; $(2,4): \frac{2}{9}$ (remaining cells 0) | M1, A2, [3] | Consider at least 6 $(U,V)$ pairs; all correct. A1 for at least 9 correct cells |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \frac{8}{9}$ or $E(Y) = 4$ | B1 | |
| $E(XY) = \frac{32}{9}$ | B1 | CWO |
| $\text{Cov}(X,Y) = \frac{32}{9} - 4 \times \frac{8}{9} = 0$ | M1A1, [4] | |
## Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Consider e.g. $P(X=1) \times P(Y=2) = \frac{4}{9} \times \frac{1}{9}$ | M1 | SC B1 no, from incorrect covariance |
| $\neq P(X=1, Y=2) = 0$ so no | A1, [2] | |
## Question 5(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{2}{9} \div \frac{4}{9} = \frac{1}{2}$ | M1A1, [2] | |
---5 The independent discrete random variables $U$ and $V$ can each take the values 1, 2 and 3, all with probability $\frac { 1 } { 3 }$. The random variables $X$ and $Y$ are defined as follows:
$$X = | U - V | , Y = U + V .$$
(i) In the Printed Answer Book complete the table showing the joint probability distribution of $X$ and $Y$.\\
(ii) Find $\operatorname { Cov } ( X , Y )$.\\
(iii) State with a reason whether $X$ and $Y$ are independent.\\
(iv) Find $\mathrm { P } ( Y = 3 \mid X = 1 )$.
\hfill \mbox{\textit{OCR S4 2018 Q5 [11]}}