| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Sample mean distribution of Poisson |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution properties (sum of independent Poissons, conditional probability using Bayes' theorem, and CLT for sample means). Part (a) requires summing probabilities for Po(4.1), part (b) applies conditional probability formula, and part (c) uses normal approximation. All are standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 2.03d Calculate conditional probability: from first principles5.02i 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda = 1.9 + 2.2\ [= 4.1]\) | B1 | |
| \(e^{-4.1}\left(1 + 4.1 + \frac{4.1^2}{2!} + \frac{4.1^3}{3!}\right)\) or \(e^{-4.1}(1 + 4.1 + 8.405 + 11.487)\) or \(0.01657 + 0.06795 + 0.13929 + 0.19037\) | M1 | Allow any \(\lambda\). Allow one end error. Must see expression. |
| \(= 0.414\) (3sf) | A1 | SC: unsupported answer 0.414 scores B1 B1 |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X + Y < 4 \text{ and } X = 2) = P(2,\ 0\ \text{or}\ 2,\ 1)\) | M1 | Stated or implied |
| \(= e^{-1.9}\times\frac{1.9^2}{2}\left(e^{-2.2} + e^{-2.2}\times2.2\right)\) \([= 0.0957]\) | M1 | |
| \(P(X = 2 \mid X + Y < 4) = \frac{0.0957}{0.414}\) | M1 | Attempt \(\frac{P(X+Y<4 \text{ and } X=2)}{P(X+Y<4)}\). Prob for denominator can be found in (a). |
| \(0.231\) (3sf) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X+Y) = 4.1\), \(\text{Var}(X+Y) = 4.1\) or \(\text{Po}(246)\) | B1 | SOI |
| Normal and \(\text{var} = \frac{4.1}{60}\), or normal and \(\text{var} = 246\) | M1 | |
| \(\frac{4.0 - 4.1}{\sqrt{4.1 \div 60}}\) or totals method \(\frac{240 - 246}{\sqrt{246}}\) or use of continuity correction | M1 | No mixed methods. Or continuity correction: \(\frac{4.0 - \frac{1}{120} - 4.1}{\sqrt{(4.1 \div 60)}}\) or \(\frac{239.5 - 246}{\sqrt{246}}\). Condone incorrect continuity correction for M1. |
| \(= -0.383\) (3sf) | A1 | \(= -0.414\) |
| \(\Phi(-0.383) = 1 - \Phi(0.383)\) | M1 | \(\Phi(-0.414) = 1 - \Phi(0.414)\) |
| \(= 0.351\) (3sf) | A1 | \(= 0.340\) or \(0.339\) |
| [6] |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = 1.9 + 2.2\ [= 4.1]$ | B1 | |
| $e^{-4.1}\left(1 + 4.1 + \frac{4.1^2}{2!} + \frac{4.1^3}{3!}\right)$ or $e^{-4.1}(1 + 4.1 + 8.405 + 11.487)$ or $0.01657 + 0.06795 + 0.13929 + 0.19037$ | M1 | Allow any $\lambda$. Allow one end error. Must see expression. |
| $= 0.414$ (3sf) | A1 | SC: unsupported answer 0.414 scores **B1 B1** |
| **Total: 3** | | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X + Y < 4 \text{ and } X = 2) = P(2,\ 0\ \text{or}\ 2,\ 1)$ | M1 | Stated or implied |
| $= e^{-1.9}\times\frac{1.9^2}{2}\left(e^{-2.2} + e^{-2.2}\times2.2\right)$ $[= 0.0957]$ | M1 | |
| $P(X = 2 \mid X + Y < 4) = \frac{0.0957}{0.414}$ | M1 | Attempt $\frac{P(X+Y<4 \text{ and } X=2)}{P(X+Y<4)}$. Prob for denominator can be found in (a). |
| $0.231$ (3sf) | A1 | |
| **Total: 4** | | |
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X+Y) = 4.1$, $\text{Var}(X+Y) = 4.1$ or $\text{Po}(246)$ | **B1** | SOI |
| Normal and $\text{var} = \frac{4.1}{60}$, or normal and $\text{var} = 246$ | **M1** | |
| $\frac{4.0 - 4.1}{\sqrt{4.1 \div 60}}$ or totals method $\frac{240 - 246}{\sqrt{246}}$ or use of continuity correction | **M1** | No mixed methods. Or continuity correction: $\frac{4.0 - \frac{1}{120} - 4.1}{\sqrt{(4.1 \div 60)}}$ or $\frac{239.5 - 246}{\sqrt{246}}$. Condone incorrect continuity correction for M1. |
| $= -0.383$ (3sf) | **A1** | $= -0.414$ |
| $\Phi(-0.383) = 1 - \Phi(0.383)$ | **M1** | $\Phi(-0.414) = 1 - \Phi(0.414)$ |
| $= 0.351$ (3sf) | **A1** | $= 0.340$ or $0.339$ |
| | **[6]** | |7 The independent random variables $X$ and $Y$ have the distributions $\operatorname { Po } ( 1.9 )$ and $\operatorname { Po } ( 2.2 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X + Y < 4 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{7c078a14-98f9-4292-ae76-a2642238176f-10_74_1581_406_322}\\
\includegraphics[max width=\textwidth, alt={}, center]{7c078a14-98f9-4292-ae76-a2642238176f-10_75_1581_497_322}
\item Find the probability that $X = 2$ given that $X + Y < 4$.\\
\includegraphics[max width=\textwidth, alt={}, center]{7c078a14-98f9-4292-ae76-a2642238176f-10_2715_35_144_2012}
\item A sample of 60 randomly chosen pairs of values of $X$ and $Y$ is taken,and the value of $X + Y$ is calculated for each pair.The sample mean of these 60 values is found.
Find the probability that the sample mean of $X + Y$ is less than 4.0 .\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q7 [13]}}