| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Joint distribution with covariance calculation |
| Difficulty | Standard +0.3 This is a straightforward S4 joint distribution question requiring standard calculations: marginal distributions, E(X), E(Y), E(XY) for covariance, independence check via P(X∩Y)=P(X)P(Y), and conditional probability. All techniques are routine for S4 students with no novel insight required, making it slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \cline { 2 - 5 } \multicolumn{2}{c|}{} | \(Y\) | |||
| \cline { 2 - 5 } \multicolumn{2}{c|}{} | 0 | 1 | 2 | |
| \multirow{3}{*}{\(X\)} | 0 | 0.07 | 0.07 | 0.16 |
| \cline { 2 - 5 } | 1 | 0.06 | 0.09 | 0.15 |
| \cline { 2 - 5 } | 2 | 0.07 | 0.14 | 0.19 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = 1.1\) | B1 | |
| \(E(Y) = 1.3\) | B1 | |
| \(E(XY) = 1.43\) | B1 | |
| \(\text{Cov}(XY) = \text{"1.43"} - \text{"1.1"} \times \text{"1.3"}\) | M1 | |
| \(0\) | A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| e.g. \(P(X=0) \times P(Y=1) \neq P(X=0, Y=1)\) | M1 | Or conditional probs. Consider any of \((0,y),(2,y)\) |
| Not independent | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{0.15}{(0.15 + 0.14)}\) | M1 | |
| \(0.517\) | A1 | Allow \(\frac{15}{29}\) |
| [2] |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = 1.1$ | B1 | |
| $E(Y) = 1.3$ | B1 | |
| $E(XY) = 1.43$ | B1 | |
| $\text{Cov}(XY) = \text{"1.43"} - \text{"1.1"} \times \text{"1.3"}$ | M1 | |
| $0$ | A1 | |
| **[5]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. $P(X=0) \times P(Y=1) \neq P(X=0, Y=1)$ | M1 | Or conditional probs. Consider any of $(0,y),(2,y)$ |
| Not independent | A1 | |
| **[2]** | | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{0.15}{(0.15 + 0.14)}$ | M1 | |
| $0.517$ | A1 | Allow $\frac{15}{29}$ |
| **[2]** | | |
---3 The table shows the joint probability distribution of two random variables $X$ and $Y$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\cline { 2 - 5 }
\multicolumn{2}{c|}{} & \multicolumn{3}{|c|}{$Y$} \\
\cline { 2 - 5 }
\multicolumn{2}{c|}{} & 0 & 1 & 2 \\
\hline
\multirow{3}{*}{$X$} & 0 & 0.07 & 0.07 & 0.16 \\
\cline { 2 - 5 }
& 1 & 0.06 & 0.09 & 0.15 \\
\cline { 2 - 5 }
& 2 & 0.07 & 0.14 & 0.19 \\
\hline
\end{tabular}
\end{center}
(i) Find $\operatorname { Cov } ( X , Y )$.\\
(ii) Are $X$ and $Y$ independent? Give a reason for your answer.\\
(iii) Find $\mathrm { P } ( X = 1 \mid X Y = 2 )$.
\hfill \mbox{\textit{OCR S4 2016 Q3 [9]}}