| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2010 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Critical region or test statistic properties |
| Difficulty | Standard +0.3 This question involves standard PGF differentiation techniques (finding E(X) and P(X=2)) and basic Wilcoxon rank-sum test setup. The PGF work is routine A-level Further Maths material requiring straightforward application of formulas, while the Wilcoxon part appears to be just defining the test statistic rather than performing complex calculations. Slightly easier than average due to being mostly procedural. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(G'(t) = 8te^{4t^2}/e^4\) | M1A1 | M1 for \(ct^2/e^4\) |
| \(E(X) = G'(1) = 8\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER: \(G(t) = e^{-4}(1 + 4t^2 + \ldots)\) | M1A1 | Expand in powers of \(t\) |
| \(P(X=2) = \) coefficient of \(t^2 = 4e^{-4}\) or \(4/e^4\) or \(0.0733\) | A1 [3] | |
| OR: \(G''(t) = (8+64t^2)e^{4t^2-4}\) | M1A1 | M1 for reasonable attempt at \(M''(t)\) |
| \(P(X=2) = \frac{1}{2}G''(0) = 4e^{-4}\) or \(4/e^4\) or \(0.0733\) | A1 [6] |
## Question 2:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $G'(t) = 8te^{4t^2}/e^4$ | M1A1 | M1 for $ct^2/e^4$ |
| $E(X) = G'(1) = 8$ | A1 **[3]** | |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| EITHER: $G(t) = e^{-4}(1 + 4t^2 + \ldots)$ | M1A1 | Expand in powers of $t$ |
| $P(X=2) = $ coefficient of $t^2 = 4e^{-4}$ or $4/e^4$ or $0.0733$ | A1 **[3]** | |
| OR: $G''(t) = (8+64t^2)e^{4t^2-4}$ | M1A1 | M1 for reasonable attempt at $M''(t)$ |
| $P(X=2) = \frac{1}{2}G''(0) = 4e^{-4}$ or $4/e^4$ or $0.0733$ | A1 **[6]** | |
---2 The probability generating function of the discrete random variable $X$ is $\frac { \mathrm { e } ^ { 4 t ^ { 2 } } } { \mathrm { e } ^ { 4 } }$. Find\\
(i) $\mathrm { E } ( X )$,\\
(ii) $\mathrm { P } ( X = 2 )$.\\
$3 X _ { 1 }$ and $X _ { 2 }$ are continuous random variables. Random samples of 5 observations of $X _ { 1 }$ and 6 observations of $X _ { 2 }$ are taken. No two observations are equal. The 11 observations are ranked, lowest first, and the sum of the ranks of the observations of $X _ { 1 }$ is denoted by $R$.\\
\hfill \mbox{\textit{OCR S4 2010 Q2 [6]}}