| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Finding minimum n for P(X=0) threshold |
| Difficulty | Standard +0.3 Part (i) is a standard Poisson approximation to binomial with straightforward calculation of P(X>3). Part (ii) requires solving P(X=0) = e^(-λ) < 0.05 for n, involving logarithms and rounding up—slightly above routine but still a textbook exercise with clear method once the Poisson parameter λ=n/2500 is identified. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Po}(1.6)\) stated or implied | M1 | |
| \(P(X > 3) = 1 - e^{-1.6}\left(1 + 1.6 + \frac{1.6^2}{2} + \frac{1.6^3}{3!}\right)\) | M1 | Allow M1 for \(1 - P(X \leqslant 3)\), incorrect \(\lambda\) and allow one end error |
| \(= 0.0788\) (3 sf) | A1 | SR Use of Bin scores B1 only for 0.0788 |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = \dfrac{n}{2500}\) | B1 | Alt method 1: \(e^{-\mu} < 0.05\); Alt method 2: \(\dfrac{2499}{2500}\) |
| \(e^{-\frac{n}{2500}} < 0.05\); Allow incorrect \(\lambda\) | M1 | Alt method 2: \(\left(\dfrac{2499}{2500}\right)^n < 0.05\) |
| \(-\dfrac{n}{2500} < \ln 0.05\); Attempt ln both sides; \(n > 7489.3\) (1 dp) | M1 | Alt method 1: \(-\mu < \ln 0.05\) \((\mu > 2.9957)\); Alt method 2: \(n\ln\dfrac{2499}{2500} < \ln 0.05\) |
| Smallest \(n = 7490\) | A1 | Alt method 1: \(n = \mu \times 2500\), Smallest \(n = 7490\); Alt method 2: Smallest \(n = 7488\) |
| Total: 4 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Po}(1.6)$ stated or implied | M1 | |
| $P(X > 3) = 1 - e^{-1.6}\left(1 + 1.6 + \frac{1.6^2}{2} + \frac{1.6^3}{3!}\right)$ | M1 | Allow M1 for $1 - P(X \leqslant 3)$, incorrect $\lambda$ and allow one end error |
| $= 0.0788$ (3 sf) | A1 | SR Use of Bin scores B1 only for 0.0788 |
| **Total** | **3** | |
## Question 5(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = \dfrac{n}{2500}$ | B1 | Alt method 1: $e^{-\mu} < 0.05$; Alt method 2: $\dfrac{2499}{2500}$ |
| $e^{-\frac{n}{2500}} < 0.05$; Allow incorrect $\lambda$ | M1 | Alt method 2: $\left(\dfrac{2499}{2500}\right)^n < 0.05$ |
| $-\dfrac{n}{2500} < \ln 0.05$; Attempt ln both sides; $n > 7489.3$ (1 dp) | M1 | Alt method 1: $-\mu < \ln 0.05$ $(\mu > 2.9957)$; Alt method 2: $n\ln\dfrac{2499}{2500} < \ln 0.05$ |
| Smallest $n = 7490$ | A1 | Alt method 1: $n = \mu \times 2500$, Smallest $n = 7490$; Alt method 2: Smallest $n = 7488$ |
| **Total: 4** | | |
---5 On average, 1 in 2500 adults has a certain medical condition.\\
(i) Use a suitable approximation to find the probability that, in a random sample of 4000 people, more than 3 have this condition.\\
(ii) In a random sample of $n$ people, where $n$ is large, the probability that none has the condition is less than 0.05 . Find the smallest possible value of $n$.\\
\hfill \mbox{\textit{CAIE S2 Q5 [7]}}