| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Standard +0.3 Part (i) requires recall of standard Poisson assumptions (routine bookwork). Parts (ii) and (iii) involve straightforward Poisson probability calculations and a normal approximation with parameter scaling—all standard S2 techniques with no novel problem-solving required. Slightly above average due to the multi-part structure and need to adjust the Poisson parameter for different time periods. |
| Spec | 2.04d Normal approximation to binomial2.05a Hypothesis testing language: null, alternative, p-value, significance5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Patients arrive at constant mean rate | B1 | B1 For first correct |
| Patients arrive at random | B1 | B1 For second correct |
| Patients arrive independently | Must be in context | |
| Patients arrive singly | SR B1 For two correct but not in context | |
| [2] | ||
| (ii) (a) \(1 - e^{-4.2} = 0.985\) | M1 A1 [2] | Correct expression |
| (ii) (b) \(4.2 \times \frac{10}{15}\) oe | B1 | |
| \(e^{-2.8} \times (1 + 2.8 + \frac{2.8^2}{2!} + \frac{2.8^3}{3!}) = 0.692\) | M1 A1 [3] | Allow extra term \(e^{-2.8} \times \frac{2.8^4}{4!}\); Allow incorrect \(\lambda\) (not 4.2) |
| (iii) N(336, 336) stated or implied | B1 | |
| \(\frac{370.5 - 336}{\sqrt{336}} (= 1.882)\) | M1 | ft "336"Allow wrong or no cc or no \(\sqrt{}\) |
| \(1 - \Phi(\text{"-1.882"}) = 0.0300/0.0299\) | M1 A1 [4] | Standardising with correct cc and no \(\sqrt{}\); Allow 0.03 |
**(i)** Patients arrive at constant mean rate | B1 | B1 For first correct
Patients arrive at random | B1 | B1 For second correct
Patients arrive independently | | Must be in context
Patients arrive singly | | SR B1 For two correct but not in context
| [2] |
**(ii) (a)** $1 - e^{-4.2} = 0.985$ | M1 A1 [2] | Correct expression
**(ii) (b)** $4.2 \times \frac{10}{15}$ oe | B1 |
$e^{-2.8} \times (1 + 2.8 + \frac{2.8^2}{2!} + \frac{2.8^3}{3!}) = 0.692$ | M1 A1 [3] | Allow extra term $e^{-2.8} \times \frac{2.8^4}{4!}$; Allow incorrect $\lambda$ (not 4.2)
**(iii)** N(336, 336) stated or implied | B1 |
$\frac{370.5 - 336}{\sqrt{336}} (= 1.882)$ | M1 | ft "336"Allow wrong or no cc or no $\sqrt{}$
$1 - \Phi(\text{"-1.882"}) = 0.0300/0.0299$ | M1 A1 [4] | Standardising with correct cc and no $\sqrt{}$; Allow 0.037 A clinic deals only with flu vaccinations. The number of patients arriving every 15 minutes is modelled by the random variable $X$ with distribution $\operatorname { Po } ( 4.2 )$.\\
(i) State two assumptions required for the Poisson model to be valid.\\
(ii) Find the probability that
\begin{enumerate}[label=(\alph*)]
\item at least 1 patient will arrive in a 15-minute period,
\item fewer than 4 patients will arrive in a 10-minute period.\\
(iii) The clinic is open for 20 hours each week. At the beginning of one week the clinic has enough vaccine for 370 patients. Use a suitable approximation to find the probability that this will not be enough vaccine for that week.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2010 Q7 [11]}}