| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | November |
| 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 e^(-n/2500) < 0.05 using logarithms, which is routine manipulation. Both parts follow textbook procedures with no novel insight required, making this slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x! |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| 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 [3] | SR Use of Bin scores B1 only for 0.0788 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda = \frac{n}{2500}\) | B1 | or \(e^{-\mu} < 0.05\) M1; or \(\frac{2499}{2500}\) B1 |
| \(e^{-\frac{n}{2500}} < 0.05\) Allow incorrect \(\lambda\) | M1 | \(\left(\frac{2499}{2500}\right)^n < 0.05\) M1 |
| \(-\frac{n}{2500} < \ln 0.05\) Attempt ln both sides | M1 | \(-\mu < \ln 0.05\) M1 \((\mu > 2.9957)\); \(n\ln\frac{2499}{2500} < \ln 0.05\) M1 |
| \(n > 7489.3\) (1 dp); Smallest \(n = 7490\) | A1 [4] | \(n = \mu \times 2500\) B1; Smallest \(n = 7490\) A1 |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| 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 [3] | SR Use of Bin scores B1 only for 0.0788 |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = \frac{n}{2500}$ | B1 | or $e^{-\mu} < 0.05$ M1; or $\frac{2499}{2500}$ B1 |
| $e^{-\frac{n}{2500}} < 0.05$ Allow incorrect $\lambda$ | M1 | $\left(\frac{2499}{2500}\right)^n < 0.05$ M1 |
| $-\frac{n}{2500} < \ln 0.05$ Attempt ln both sides | M1 | $-\mu < \ln 0.05$ M1 $(\mu > 2.9957)$; $n\ln\frac{2499}{2500} < \ln 0.05$ M1 |
| $n > 7489.3$ (1 dp); Smallest $n = 7490$ | A1 [4] | $n = \mu \times 2500$ B1; Smallest $n = 7490$ A1 |
---5 On average, 1 in 2500 adults has a certain medical condition.\\
\begin{enumerate}[label=(\roman*)]
\item Use a suitable approximation to find the probability that, in a random sample of 4000 people, more than 3 have this condition.
\item 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 $n$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2015 Q5 [7]}}