| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Joint distribution with independence testing |
| Difficulty | Standard +0.3 This is a standard S4 joint distribution question requiring routine calculations: finding marginal expectations/variances, testing independence via P(X,Y)=P(X)P(Y), applying covariance and variance formulas, and conditional probability. All techniques are textbook applications with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Either: Show correct process for comp'n | M1 | correct way round and in terms of \(x\) |
| Obtain \(y = 3(3x + 7) - 2\) | A1 | or equiv |
| Obtain \(x = -\frac{19}{9}\) | A1 3 | or exact equiv; condone absence of \(y = 0\) |
| Or: Use \(fg(x) = 0\) to obtain \(g(x) = \frac{2}{3}\) | B1 | |
| Attempt solution of \(g(x) = \frac{2}{3}\) | M1 | |
| Obtain \(x = -\frac{19}{9}\) | A1 (3) | or exact equiv; condone absence of \(y = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt formation of one of the equations | M1 | \(3x + 7 = \frac{x-7}{3}\) or \(3x + 7 = x\) or \(\frac{x-7}{3} = x\) |
| Obtain \(x = -\frac{2}{3}\) | A1 | or equiv |
| Obtain \(y = -\frac{2}{3}\) | A1√ 3 | or equiv; following their value of \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt solution of modulus equation | M1 | squaring both sides to obtain 3-term quadratics or forming linear equation with signs of 3x different on each side |
| Obtain \(-12x + 4 = 42x + 49\) or \(3x - 2 = -3x - 7\) | A1 | or equiv |
| Obtain \(x = -\frac{5}{6}\) | A1 | or exact equiv; as final answer |
| Obtain \(y = \frac{2}{3}\) | A1 4 | or equiv; and no other pair of answers |
### (i)
Either: Show correct process for comp'n | M1 | correct way round and in terms of $x$ |
Obtain $y = 3(3x + 7) - 2$ | A1 | or equiv |
Obtain $x = -\frac{19}{9}$ | A1 3 | or exact equiv; condone absence of $y = 0$ |
Or: Use $fg(x) = 0$ to obtain $g(x) = \frac{2}{3}$ | B1 |
Attempt solution of $g(x) = \frac{2}{3}$ | M1 |
Obtain $x = -\frac{19}{9}$ | A1 (3) | or exact equiv; condone absence of $y = 0$ |
### (ii)
Attempt formation of one of the equations | M1 | $3x + 7 = \frac{x-7}{3}$ or $3x + 7 = x$ or $\frac{x-7}{3} = x$ |
Obtain $x = -\frac{2}{3}$ | A1 | or equiv |
Obtain $y = -\frac{2}{3}$ | A1√ 3 | or equiv; following their value of $x$ |
### (iii)
Attempt solution of modulus equation | M1 | squaring both sides to obtain 3-term quadratics or forming linear equation with signs of 3x different on each side |
Obtain $-12x + 4 = 42x + 49$ or $3x - 2 = -3x - 7$ | A1 | or equiv |
Obtain $x = -\frac{5}{6}$ | A1 | or exact equiv; as final answer |
Obtain $y = \frac{2}{3}$ | A1 4 | or equiv; and no other pair of answers |5 Alana and Ben work for an estate agent. The joint probability distribution of the number of houses they sell in a randomly chosen week, $X _ { A }$ and $X _ { B }$ respectively, is shown in the table.\\
\includegraphics[max width=\textwidth, alt={}, center]{f1879b0f-17e3-41b4-af38-a843b67c5301-3_405_602_370_781}\\
(i) Find $\mathrm { E } \left( X _ { A } \right)$ and $\operatorname { Var } \left( X _ { A } \right)$.\\
(ii) Determine whether $X _ { A }$ and $X _ { B }$ are independent.\\
(iii) Given that $\mathrm { E } \left( X _ { B } \right) = 1.15 , \operatorname { Var } \left( X _ { B } \right) = 0.8275$ and $\mathrm { E } \left( X _ { A } X _ { B } \right) = 1.09$, find $\operatorname { Cov } \left( X _ { A } , X _ { B } \right)$ and $\operatorname { Var } \left( X _ { A } - X _ { B } \right)$.\\
(iv) During a particular week only one house was sold by Alana and Ben. Find the probability that it was sold by Alana.
\hfill \mbox{\textit{OCR S4 2009 Q5 [13]}}