| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2002 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Sample mean distribution of Poisson |
| Difficulty | Standard +0.3 This question tests standard Poisson distribution properties: (i) uses the fact that sum of independent Poissons is Poisson with combined mean, requiring straightforward probability calculation; (ii) applies CLT to find probability about sample mean. Both parts are routine applications of well-known results with no conceptual challenges, though (ii) requires recognizing when to use normal approximation. Slightly above average difficulty due to being Further Maths content and requiring multiple standard techniques. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02n Sum of Poisson variables: is Poisson5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) new mean \(= 5.6\) | B1 | For new mean. |
| \(P(X+Y > 3) = 1 - \{P(0) + P(1) + P(2) + P(3)\}\) | M1 | For evaluating \(1 -\) some Poisson probabilities. |
| \(= 1 - e^{-5.6}\left(1 + 5.6 + \frac{5.6^2}{2!} + \frac{5.6^2}{3!}\right) = 1 - 0.809 = 0.809\) | A1, A1 (4 marks) | For correct expression. For correct answer. |
| (ii) \(\bar{X} \sim N\left(2.5, \frac{2.5}{80}\right)\) or equiv. method using totals \(N(200, 200)\) | M1, A1 | For using normal distribution with mean \(2.5/200\). For correct variance. |
| \(P(X < 2.4) = \Phi\left(\frac{2.4 - 2.5}{\sqrt{2.5/80}}\right)\) or \(\Phi\left(\frac{192-200}{\sqrt{200}}\right)\) | M1 | For standardising and using normal tables. |
| \(= \Phi(-0.566) = 1 - 0.7143 = 0.286\) | A1 (4 marks) | For correct answer. |
(i) new mean $= 5.6$ | B1 | For new mean.
$P(X+Y > 3) = 1 - \{P(0) + P(1) + P(2) + P(3)\}$ | M1 | For evaluating $1 -$ some Poisson probabilities.
$= 1 - e^{-5.6}\left(1 + 5.6 + \frac{5.6^2}{2!} + \frac{5.6^2}{3!}\right) = 1 - 0.809 = 0.809$ | A1, A1 (4 marks) | For correct expression. For correct answer.
(ii) $\bar{X} \sim N\left(2.5, \frac{2.5}{80}\right)$ or equiv. method using totals $N(200, 200)$ | M1, A1 | For using normal distribution with mean $2.5/200$. For correct variance.
$P(X < 2.4) = \Phi\left(\frac{2.4 - 2.5}{\sqrt{2.5/80}}\right)$ or $\Phi\left(\frac{192-200}{\sqrt{200}}\right)$ | M1 | For standardising and using normal tables.
$= \Phi(-0.566) = 1 - 0.7143 = 0.286$ | A1 (4 marks) | For correct answer.
---$X$ and $Y$ are independent random variables each having a Poisson distribution. $X$ has mean 2.5 and $Y$ has mean 3.1.
\begin{enumerate}[label=(\roman*)]
\item Find P$(X + Y > 3)$. [4]
\item A random sample of 80 values of $X$ is taken. Find the probability that the sample mean is less than 2.4. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2002 Q5 [8]}}