| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | Combined independent Poisson probabilities |
| Difficulty | Standard +0.8 This question combines Poisson approximation to binomial with independence (part a) and algebraic manipulation of Poisson probability functions involving two parameters (part b). Part (a) requires recognizing when approximation is valid, computing λ values, and handling combined independent events. Part (b) involves setting up equations from Poisson pmf expressions and solving for k through substitution—requiring careful algebraic manipulation beyond routine application. The multi-step reasoning and parameter relationships elevate this above standard single-distribution problems. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.024 \times 50 [= 1.2]\) and \(0.018 \times 60 [= 1.08]\) | B1 | |
| \((1 - e^{-1.2}(1 + 1.2)) \times (1 - e^{-1.08}(1 + 1.08))\) | M1 | For \((1 - e^{-\lambda}(1+\lambda)) \times (1 - e^{-\mu}(1+\mu))\) any \(\lambda\), \(\mu\) \((\lambda \neq \mu)\). Allow one end error on either or both terms |
| \(= 0.0991\) (3 sf) | A1 | Unsupported answer scores maximum SC B1 B1. SC Use of binomial 0.0994 scores B1 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = 0.024 \times 50 + 0.018 \times 60\) | M1 | or *their* \(1.2 + 1.08\) (NB \(0.024 + 0.018\) is M0) |
| \(1 - e^{-2.28} \times \left(1 + 2.28 + \frac{2.28^2}{2!} + \frac{2.28^3}{3!}\right)\) | M1 | any \(\lambda\) and allow one end error |
| \(= 0.197\) (3 sf) | A1 | Unsupported answer scores maximum SC B2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-\lambda} = \left[e^{-\mu}\right]^2 = e^{-2\mu}\) | M1 | |
| \(e^{-\lambda} \times \frac{\lambda^2}{2} = k\left[e^{-\mu} \times \mu\right]^2\) | M1 | |
| \(e^{-2\mu} \times 2\mu^2 = k \times e^{-2\mu} \times \mu^2\) | M1 | OE. Use of \(\lambda = 2\mu\) to find equation in \(\mu\) and \(k\) only (or \(\lambda\) and \(k\) only) |
| \(k = 2\) | A1 |
## Question 7:
### Part 7(a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.024 \times 50 [= 1.2]$ and $0.018 \times 60 [= 1.08]$ | B1 | |
| $(1 - e^{-1.2}(1 + 1.2)) \times (1 - e^{-1.08}(1 + 1.08))$ | M1 | For $(1 - e^{-\lambda}(1+\lambda)) \times (1 - e^{-\mu}(1+\mu))$ any $\lambda$, $\mu$ $(\lambda \neq \mu)$. Allow one end error on either or both terms |
| $= 0.0991$ (3 sf) | A1 | Unsupported answer scores maximum SC B1 B1. SC Use of binomial 0.0994 scores B1 only |
### Part 7(a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 0.024 \times 50 + 0.018 \times 60$ | M1 | or *their* $1.2 + 1.08$ (NB $0.024 + 0.018$ is M0) |
| $1 - e^{-2.28} \times \left(1 + 2.28 + \frac{2.28^2}{2!} + \frac{2.28^3}{3!}\right)$ | M1 | any $\lambda$ and allow one end error |
| $= 0.197$ (3 sf) | A1 | Unsupported answer scores maximum SC B2 |
### Part 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-\lambda} = \left[e^{-\mu}\right]^2 = e^{-2\mu}$ | M1 | |
| $e^{-\lambda} \times \frac{\lambda^2}{2} = k\left[e^{-\mu} \times \mu\right]^2$ | M1 | |
| $e^{-2\mu} \times 2\mu^2 = k \times e^{-2\mu} \times \mu^2$ | M1 | OE. Use of $\lambda = 2\mu$ to find equation in $\mu$ and $k$ only (or $\lambda$ and $k$ only) |
| $k = 2$ | A1 | |7
\begin{enumerate}[label=(\alph*)]
\item Two ponds, $A$ and $B$, each contain a large number of fish. It is known that $2.4 \%$ of fish in pond $A$ are carp and $1.8 \%$ of fish in pond $B$ are carp. Random samples of 50 fish from pond $A$ and 60 fish from pond $B$ are selected.
Use appropriate Poisson approximations to find the following probabilities.
\begin{enumerate}[label=(\roman*)]
\item The samples contain at least 2 carp from pond $A$ and at least 2 carp from pond $B$.
\item The samples contain at least 4 carp altogether.
\end{enumerate}\item The random variables $X$ and $Y$ have the distributions $\operatorname { Po } ( \lambda )$ and $\operatorname { Po } ( \mu )$ respectively. It is given that
\begin{itemize}
\item $\mathrm { P } ( X = 0 ) = [ \mathrm { P } ( Y = 0 ) ] ^ { 2 }$,
\item $\mathrm { P } ( X = 2 ) = k [ \mathrm { P } ( Y = 1 ) ] ^ { 2 }$, where $k$ is a non-zero constant.
\end{itemize}
Find the value of $k$.\\
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 2022 Q7 [10]}}