| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Joint distribution with covariance calculation |
| Difficulty | Standard +0.8 This S4 joint distribution question requires systematic enumeration of 9 probability values, finding marginal distributions, computing E(X), E(Y), E(XY), and Cov(X,Y) with multiple summations. While methodical rather than conceptually deep, the computational load and multi-step covariance calculation place it moderately above average difficulty for A-level, though standard for Further Maths S4. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a, 2a, 3a;\; 2a, 3a, 4a;\; 3a, 4a, 5a\) | B1 | Allow \(a(0+0+1)\), etc |
| \(a = \tfrac{1}{27}\) AG | B1 | Must see \((27a)\text{oe}=1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x\): \(0, 1, 2\); \(p\): \(\frac{2}{9}, \frac{3}{9}, \frac{4}{9}\) | B1 | oe |
| \(E(X) = \tfrac{11}{9}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(xy\): \(0,1,2,4\); \(p\): \(\frac{11}{27}, \frac{3}{27}, \frac{8}{27}, \frac{5}{27}\) | B1 | Seen or implied e.g. by \(3a+8a+8a+20a\) |
| \(E(XY) = \tfrac{39}{27}\), \(E(Y) = \tfrac{11}{9}\) ft | B1, B1ft | \(E(XY) = \frac{13}{9}\) |
| Use \(\text{Cov}(XY) = E(XY) - E(X)E(Y)\) | M1 | |
| \(-\tfrac{4}{81}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Cov} \neq 0\) | M1 | ft non-zero Cov. OR e.g. \(P(0,0)\neq P(0){\times}P(0)\) |
| No | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tfrac{4}{27} \div \tfrac{12}{27}\) | M1 | \(4a/12a\) |
| \(= \tfrac{1}{3}\) | A1 |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a, 2a, 3a;\; 2a, 3a, 4a;\; 3a, 4a, 5a$ | B1 | Allow $a(0+0+1)$, etc |
| $a = \tfrac{1}{27}$ AG | B1 | Must see $(27a)\text{oe}=1$ |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x$: $0, 1, 2$; $p$: $\frac{2}{9}, \frac{3}{9}, \frac{4}{9}$ | B1 | oe |
| $E(X) = \tfrac{11}{9}$ | B1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $xy$: $0,1,2,4$; $p$: $\frac{11}{27}, \frac{3}{27}, \frac{8}{27}, \frac{5}{27}$ | B1 | Seen or implied e.g. by $3a+8a+8a+20a$ |
| $E(XY) = \tfrac{39}{27}$, $E(Y) = \tfrac{11}{9}$ ft | B1, B1ft | $E(XY) = \frac{13}{9}$ |
| Use $\text{Cov}(XY) = E(XY) - E(X)E(Y)$ | M1 | |
| $-\tfrac{4}{81}$ | A1 | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Cov} \neq 0$ | M1 | ft non-zero Cov. OR e.g. $P(0,0)\neq P(0){\times}P(0)$ |
| No | A1 | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tfrac{4}{27} \div \tfrac{12}{27}$ | M1 | $4a/12a$ |
| $= \tfrac{1}{3}$ | A1 | |
---5 Two discrete random variables $X$ and $Y$ have a joint probability distribution defined by
$$\mathrm { P } ( X = x , Y = y ) = a ( x + y + 1 ) \quad \text { for } x = 0,1,2 \text { and } y = 0,1,2 ,$$
where $a$ is a constant.\\
(i) Show that $a = \frac { 1 } { 27 }$.\\
(ii) Find $\mathrm { E } ( X )$.\\
(iii) Find $\operatorname { Cov } ( X , Y )$.\\
(iv) Are $X$ and $Y$ independent? Give a reason for your answer.\\
(v) Find $\mathrm { P } ( X = 1 \mid Y = 2 )$.
\hfill \mbox{\textit{OCR S4 2014 Q5 [13]}}