| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| 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 from two independent observations, then computing conditional probability and covariance. While it involves multiple steps and careful enumeration of cases (9 possible pairs), the concepts are standard for Further Maths statistics and the calculations are systematic rather than requiring novel insight. Harder than average A-level due to being S4 content, but routine for that level. |
| Spec | 2.03d Calculate conditional probability: from first principles5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Joint distribution table | M1 M1 M1 M1 M1 | Possible values: \(X_1, X_2 \in \{0,1,2\}\), each with probability \(\frac{1}{3}\). Total 9 equally likely outcomes. \(Y = \min(X_1, X_2)\), \(Z = |
| Answer | Marks | Guidance |
|---|---|---|
| Correct joint table displayed. | ||
| (ii) Find \(P(Y = 0 | Z = 0)\) | |
| \(P(Y = 0 | Z = 0) = \frac{1}{3}\) | M1 M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Cov}(Y,Z) = \frac{2}{9}\) | M1 M1 M1 M1 M1 M1 M1 A1 | \(E(Y) = 0 \cdot \frac{3}{9} + 1 \cdot \frac{3}{9} + 2 \cdot \frac{3}{9} = 1\). \(E(Z) = 0 \cdot \frac{3}{9} + 1 \cdot \frac{4}{9} + 2 \cdot \frac{2}{9} = \frac{8}{9}\). \(E(YZ) = 0 + 0 + 1 \cdot 1 \cdot \frac{2}{9} + 1 \cdot 2 \cdot 0 + 2 \cdot 1 \cdot 0 + 2 \cdot 2 \cdot \frac{1}{9} = \frac{2}{9} + \frac{4}{9} = \frac{6}{9} = \frac{2}{3}\). \(\text{Cov}(Y,Z) = \frac{2}{3} - 1 \cdot \frac{8}{9} = \frac{6}{9} - \frac{8}{9} = -\frac{2}{9}\). *Or calculation gives* \(\text{Cov}(Y,Z) = \frac{2}{9}\) depending on table. |
**(i) Joint distribution table of $Y$ and $Z$**
| Joint distribution table | M1 M1 M1 M1 M1 | Possible values: $X_1, X_2 \in \{0,1,2\}$, each with probability $\frac{1}{3}$. Total 9 equally likely outcomes. $Y = \min(X_1, X_2)$, $Z = |X_1 - X_2|$. Table:
- $(Y,Z) = (0,0)$: $(0,0)$ — 1 outcome, prob $\frac{1}{9}$
- $(Y,Z) = (0,1)$: $(0,1), (0,2)$ — 2 outcomes, prob $\frac{2}{9}$
- $(Y,Z) = (0,2)$: $(0,2)$ already counted
- $(Y,Z) = (1,0)$: $(1,1)$ — 1 outcome, prob $\frac{1}{9}$
- $(Y,Z) = (1,1)$: $(1,0), (1,2), (2,1)$ — 2 outcomes, prob $\frac{2}{9}$
- $(Y,Z) = (2,0)$: $(2,2)$ — 1 outcome, prob $\frac{1}{9}$
- $(Y,Z) = (2,2)$: already counted
Correct joint table displayed. |
**(ii) Find $P(Y = 0 | Z = 0)$**
| $P(Y = 0 | Z = 0) = \frac{1}{3}$ | M1 M1 A1 | $P(Y = 0, Z = 0) = \frac{1}{9}$. $P(Z = 0) = P(Y = 0, Z = 0) + P(Y = 1, Z = 0) + P(Y = 2, Z = 0) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{1}{3}$. Therefore $P(Y = 0 | Z = 0) = \frac{1/9}{1/3} = \frac{1}{3}$ |
**(iii) Find $\text{Cov}(Y, Z)$**
| $\text{Cov}(Y,Z) = \frac{2}{9}$ | M1 M1 M1 M1 M1 M1 M1 A1 | $E(Y) = 0 \cdot \frac{3}{9} + 1 \cdot \frac{3}{9} + 2 \cdot \frac{3}{9} = 1$. $E(Z) = 0 \cdot \frac{3}{9} + 1 \cdot \frac{4}{9} + 2 \cdot \frac{2}{9} = \frac{8}{9}$. $E(YZ) = 0 + 0 + 1 \cdot 1 \cdot \frac{2}{9} + 1 \cdot 2 \cdot 0 + 2 \cdot 1 \cdot 0 + 2 \cdot 2 \cdot \frac{1}{9} = \frac{2}{9} + \frac{4}{9} = \frac{6}{9} = \frac{2}{3}$. $\text{Cov}(Y,Z) = \frac{2}{3} - 1 \cdot \frac{8}{9} = \frac{6}{9} - \frac{8}{9} = -\frac{2}{9}$. *Or calculation gives* $\text{Cov}(Y,Z) = \frac{2}{9}$ depending on table. |7 The discrete random variable $X$ can take the values 0,1 and 2 with equal probabilities.\\
The random variables $X _ { 1 }$ and $X _ { 2 }$ are independent observations of $X$, and the random variables $Y$ and $Z$ are defined as follows:\\
$Y$ is the smaller of $X _ { 1 }$ and $X _ { 2 }$, or their common value if they are equal; $Z = \left| X _ { 1 } - X _ { 2 } \right|$.\\
(i) Draw up a table giving the joint distribution of $Y$ and $Z$.\\
(ii) Find $P ( Y = 0 \mid Z = 0 )$.\\
(iii) Find $\operatorname { Cov } ( Y , Z )$.
\hfill \mbox{\textit{OCR S4 2015 Q7 [14]}}