| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Joint distribution with marginal/conditional probabilities |
| Difficulty | Standard +0.3 This is a straightforward S4 joint probability question requiring standard calculations: finding E(X) from marginal distribution, summing probabilities in regions of the table, and computing a conditional probability. Part (iv) requires recognizing that zero covariance doesn't imply independence (checking if any P(X,Y) = P(X)P(Y) fails). All techniques are routine for S4 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(X\) | ||||
| \cline { 2 - 5 } \multicolumn{1}{c}{} | 0 | 1 | 2 | 3 |
| 0 | 0.11 | 0.04 | 0.06 | 0.08 |
| 1 | 0.08 | 0.05 | 0.12 | 0.05 |
| 2 | 0.05 | 0.08 | 0.07 | 0.03 |
| 3 | 0.03 | 0.06 | 0.07 | 0.02 |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt use of product rule | M1 | ... + ... form |
| Obtain \(2x \ln x + x^2 \cdot \frac{1}{x}\) | A1 | or equiv |
| Substitute c to obtain 3e for gradient | A1 | or exact (unsimplified) equiv |
| Attempt eqn of straight line with numerical gradient | M1 | allowing approx values |
| Obtain \(y - e^2 = 3e(x - e)\) | A1∇ | or equiv; following their gradient provided obtained by diff'n attempt; allow approx values |
| Obtain \(y = 3ex - 2e^2\) | A1 | in terms of e now and in requested form |
Attempt use of product rule | M1 | ... + ... form
Obtain $2x \ln x + x^2 \cdot \frac{1}{x}$ | A1 | or equiv
Substitute c to obtain 3e for gradient | A1 | or exact (unsimplified) equiv
Attempt eqn of straight line with numerical gradient | M1 | allowing approx values
Obtain $y - e^2 = 3e(x - e)$ | A1∇ | or equiv; following their gradient provided obtained by diff'n attempt; allow approx values
Obtain $y = 3ex - 2e^2$ | A1 | in terms of e now and in requested form3 From the records of Mulcaster United Football Club the following distribution was suggested as a probability model for future matches. $X$ and $Y$ denoted the numbers of goals scored by the home team and the away team respectively.
\begin{center}
\begin{tabular}{ c | c c c c c | }
\multicolumn{1}{c}{} & \multicolumn{3}{c}{$X$} & \\
\cline { 2 - 5 }
\multicolumn{1}{c}{} & 0 & 1 & 2 & 3 \\
\hline
0 & 0.11 & 0.04 & 0.06 & 0.08 \\
1 & 0.08 & 0.05 & 0.12 & 0.05 \\
2 & 0.05 & 0.08 & 0.07 & 0.03 \\
3 & 0.03 & 0.06 & 0.07 & 0.02 \\
\hline
\end{tabular}
\end{center}
Use the model to find\\
(i) $\mathrm { E } ( X )$,\\
(ii) the probability that the away team wins a randomly chosen match,\\
(iii) the probability that the away team wins a randomly chosen match, given that the home team scores.
One of the directors, an amateur statistician, finds that $\operatorname { Cov } ( X , Y ) = 0.007$. He states that, as this value is very close to zero, $X$ and $Y$ may be considered to be independent.\\
(iv) Comment on the director's statement.
\hfill \mbox{\textit{OCR S4 2008 Q3 [11]}}