| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Joint probability of separate processes |
| Difficulty | Standard +0.8 This question requires understanding of Poisson distribution properties including scaling by time, independence of Poisson variables, sum of Poisson distributions, and normal approximation for large λ. Part (i) involves complementary probability calculations with scaled parameters. Part (ii) requires recognizing when to apply continuity correction and normal approximation for λ=52, which is a non-trivial conceptual step beyond routine Poisson calculations. |
| 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 Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - e^{-1.2}\) \((= 0.7534)\) | M1, A1 | M1 for Poisson either P(0 or 1) or P(0) with \(\lambda = 1.2\) or \(2.4\) or \(1.4\) or \(2.8\), accept one end error. Both expressions fully correct |
| \((1 - e^{-1.2}(1 + 1.2)) \times (1 - e^{-1.4})\) | M1 | Their Poisson P(0 or 1) × P(0) |
| \(= 0.254\) (3 s.f.) | A1 | [4] |
| (i)(b) \(\lambda = 2.6\) seen | B1 | |
| \(1 - e^{-2.6}(1 + 2.6 + 2.6^2 \div 2)\) | M1 | Poisson \(1 - \text{P}(0, 1, 2)\), allow \(1 - \text{P}(0, 1, 2, 3)\), with attempt at combined \(\lambda\) for M and W. Accept combination method: at least 4 correct terms and "1 −" M1; all terms correct B1 |
| \(= 0.482\) (3 s.f.) | A1 | [3] |
| (ii) \(\text{N}(52, 52)\) | B1 | Seen or implied |
| \(\frac{60.5 - 52}{\sqrt{52}}\) \((= 1.179)\) | M1 | Standardising with N(\(\lambda, \lambda\)) with \(\lambda = 10 \times 5.2\) or \(10 \times 2.6\). Allow with wrong or no cc or no \(\sqrt{}\) |
| \(1 - \Phi(1.179)\) | M1 | Their correct area |
| \((= 1 - 0.8808)\) | ||
| \(= 0.119\) (3 s.f.) | A1 | [4] |
**(i)(a)** $1 - e^{-1.2}(1 + 1.2)$ $(= 0.3374)$
$1 - e^{-1.2}$ $(= 0.7534)$ | M1, A1 | M1 for Poisson either P(0 or 1) or P(0) with $\lambda = 1.2$ or $2.4$ or $1.4$ or $2.8$, accept one end error. Both expressions fully correct
$(1 - e^{-1.2}(1 + 1.2)) \times (1 - e^{-1.4})$ | M1 | Their Poisson P(0 or 1) × P(0)
$= 0.254$ (3 s.f.) | A1 | [4]
**(i)(b)** $\lambda = 2.6$ seen | B1 |
$1 - e^{-2.6}(1 + 2.6 + 2.6^2 \div 2)$ | M1 | Poisson $1 - \text{P}(0, 1, 2)$, allow $1 - \text{P}(0, 1, 2, 3)$, with attempt at combined $\lambda$ for M and W. Accept combination method: at least 4 correct terms and "1 −" M1; all terms correct B1
$= 0.482$ (3 s.f.) | A1 | [3]
**(ii)** $\text{N}(52, 52)$ | B1 | Seen or implied
$\frac{60.5 - 52}{\sqrt{52}}$ $(= 1.179)$ | M1 | Standardising with N($\lambda, \lambda$) with $\lambda = 10 \times 5.2$ or $10 \times 2.6$. Allow with wrong or no cc or no $\sqrt{}$
$1 - \Phi(1.179)$ | M1 | Their correct area
$(= 1 - 0.8808)$ | |
$= 0.119$ (3 s.f.) | A1 | [4]The numbers of men and women who visit a clinic each hour are independent Poisson variables with means 2.4 and 2.8 respectively.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that, in a half-hour period,
\begin{enumerate}[label=(\alph*)]
\item 2 or more men and 1 or more women will visit the clinic, [4]
\item a total of 3 or more people will visit the clinic. [3]
\end{enumerate}
\item Find the probability that, in a 10-hour period, a total of more than 60 people will visit the clinic. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2011 Q7 [11]}}