| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson approximation with parameter finding |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson approximation to binomial with standard calculations. Part (a) requires recognizing np is small and n is large (textbook condition), parts (b-c) involve routine Poisson probability calculations, and part (d) requires solving P(Y≥1)=1-e^(-λ) which is algebraically simple. All steps are standard S2 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02b Expectation and variance: discrete random variables5.02d Binomial: mean np and variance np(1-p)5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Po}(2.5)\) | B1 | Accept Poisson with mean = 2.5 not just \(np = 2.5\) |
| \(n = 25000 > 50\) and \(np\) (or \(\lambda\)) \(= 2.5\) which is \(< 5\); or \(n = 25000 > 50\) and \(p = 0.0001 < 0.1\) | B1 | Must see 2.5 (or 0.0001) and 25000 OE, not just \(np < 5\) (or \(p < 0.1\)) and \(n > 50\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-2.5}\!\left(1 + 2.5 + \dfrac{2.5^2}{2} + \dfrac{2.5^3}{3!}\right)\) | M1 | Any \(\lambda\), accept one end error. FT binomial from part (a) scores M1 only for equivalent binomial expressions; FT normal from part (a) must use correct continuity correction and tables scores M1 only for complete method |
| \(0.758\) (3 sf) | A1 | Unsupported answer of 0.758 scores B1 instead of M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-2.5} \times \dfrac{2.5^k}{(k)!} = 2e^{-2.5} \times \dfrac{2.5^{k+1}}{(k+1)!}\) | M1 | Any \(\lambda\); FT binomial from (a) scores M1 only for equivalent binomial expression; FT from (a) normal for equivalent expressions, continuity correction must be included |
| \(k = 4\) | A1 | No errors seen; SC \(k=4\) unsupported scores B1 only, but see full Poisson expressions for P(4) and P(5) and 0.134 scores M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - e^{-\lambda} = 0.963\) | M1 | Accept *their* attempt at \(\lambda\) |
| \(\lambda = -\ln 0.037\ (= 3.2968\) or \(3.30\) or \(3.3)\) | M1 | Correct use of \(\ln\)s |
| \(n = 33000\) (3 sf) | A1 | Allow \(n = 32950\) to \(33050\) (must be an integer); SC use of binomial leading to 32967 scores B1 for \((0.9999)^n = 0.037\); B1 for 33000 to 3sf (32967) |
## Question 5:
### Part 5(a):
$\text{Po}(2.5)$ | **B1** | Accept Poisson with mean = 2.5 not just $np = 2.5$
$n = 25000 > 50$ and $np$ (or $\lambda$) $= 2.5$ which is $< 5$; or $n = 25000 > 50$ and $p = 0.0001 < 0.1$ | **B1** | Must see 2.5 (or 0.0001) and 25000 OE, not just $np < 5$ (or $p < 0.1$) and $n > 50$
### Part 5(b):
$e^{-2.5}\!\left(1 + 2.5 + \dfrac{2.5^2}{2} + \dfrac{2.5^3}{3!}\right)$ | **M1** | Any $\lambda$, accept one end error. FT binomial from part (a) scores M1 only for equivalent binomial expressions; FT normal from part (a) must use correct continuity correction and tables scores M1 only for complete method
$0.758$ (3 sf) | **A1** | Unsupported answer of 0.758 scores B1 instead of M1A1
### Part 5(c):
$e^{-2.5} \times \dfrac{2.5^k}{(k)!} = 2e^{-2.5} \times \dfrac{2.5^{k+1}}{(k+1)!}$ | **M1** | Any $\lambda$; FT binomial from (a) scores M1 only for equivalent binomial expression; FT from (a) normal for equivalent expressions, continuity correction must be included
$k = 4$ | **A1** | No errors seen; SC $k=4$ unsupported scores B1 only, but see full Poisson expressions for P(4) and P(5) and 0.134 scores M1A1
### Part 5(d):
$1 - e^{-\lambda} = 0.963$ | **M1** | Accept *their* attempt at $\lambda$
$\lambda = -\ln 0.037\ (= 3.2968$ or $3.30$ or $3.3)$ | **M1** | Correct use of $\ln$s
$n = 33000$ (3 sf) | **A1** | Allow $n = 32950$ to $33050$ (must be an integer); SC use of binomial leading to 32967 scores B1 for $(0.9999)^n = 0.037$; B1 for 33000 to 3sf (32967)
---5 Most plants of a certain type have three leaves. However, it is known that, on average, 1 in 10000 of these plants have four leaves, and plants with four leaves are called 'lucky'. The number of lucky plants in a random sample of 25000 plants is denoted by $X$.
\begin{enumerate}[label=(\alph*)]
\item State, with a justification, an approximating distribution for $X$, giving the values of any parameters.\\
Use your approximating distribution to answer parts (b) and (c).
\item Find $\mathrm { P } ( X \leqslant 3 )$.
\item Given that $\mathrm { P } ( X = k ) = 2 \mathrm { P } ( X = k + 1 )$, find $k$.\\
The number of lucky plants in a random sample of $n$ plants, where $n$ is large, is denoted by $Y$.
\item Given that $\mathrm { P } ( Y \geqslant 1 ) = 0.963$, correct to 3 significant figures, use a suitable approximating distribution to find the value of $n$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q5 [9]}}