| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| 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>2). Part (ii) requires solving P(X=0) = e^(-λ) < 0.01 using logarithms, which is a routine application of the formula but involves an extra algebraic step beyond basic recall. Overall slightly easier than average due to clear setup and standard techniques. |
| Spec | 5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda = 1.8\) | M1 | For attempt at Poisson, any mean |
| \(P(X > 2) = 1 - [P(0) + P(1) + P(2)]\) | A1 | For correct mean |
| \(= 1 - e^{-1.8}\left(1 + 1.8 + \frac{1.8^2}{2!}\right)\) | M1 | For finding \(1 - P(0) - P(1) - P(2)\) or \(1 - P(0) - P(1)\) |
| \(= 1 - 0.7306\) | ||
| \(= 0.269\) | A1 4 | For correct answer. SR1 Normal scores B1 for \(2.5 - 1.8/\sqrt{(1.7988)}\). SR2 Binomial scores M1 for complete method leading to final answer of 0.269, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda = n/1500\) | B1 | For correct Poisson mean |
| \(P(0) < 0.01\) i.e. \(e^{\frac{-n}{1500}} < 0.01\) | M1 | For equation or inequality involving their \(P(0)\) and \(0.01\) |
| \(\frac{-n}{1500} < \ln 0.01\) | ||
| \(n > 6907.7\) | ||
| \(n = 6908\) | A1 3 | For correct answer |
| OR \((1499/1500)^n < 0.01\) | (B1)(M1) | For correct Binomial \(p\). For correct equation/inequality involving their \(P(0)\) and \(0.01\) |
| \(n = 6906\) | (A1) | For correct answer |
# Question 5:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda = 1.8$ | M1 | For attempt at Poisson, any mean |
| $P(X > 2) = 1 - [P(0) + P(1) + P(2)]$ | A1 | For correct mean |
| $= 1 - e^{-1.8}\left(1 + 1.8 + \frac{1.8^2}{2!}\right)$ | M1 | For finding $1 - P(0) - P(1) - P(2)$ or $1 - P(0) - P(1)$ |
| $= 1 - 0.7306$ | | |
| $= 0.269$ | A1 **4** | For correct answer. SR1 Normal scores B1 for $2.5 - 1.8/\sqrt{(1.7988)}$. SR2 Binomial scores M1 for complete method leading to final answer of 0.269, A1 |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda = n/1500$ | B1 | For correct Poisson mean |
| $P(0) < 0.01$ i.e. $e^{\frac{-n}{1500}} < 0.01$ | M1 | For equation or inequality involving their $P(0)$ and $0.01$ |
| $\frac{-n}{1500} < \ln 0.01$ | | |
| $n > 6907.7$ | | |
| $n = 6908$ | A1 **3** | For correct answer |
| **OR** $(1499/1500)^n < 0.01$ | (B1)(M1) | For correct Binomial $p$. For correct equation/inequality involving their $P(0)$ and $0.01$ |
| $n = 6906$ | (A1) | For correct answer |
---5 Of people who wear contact lenses, 1 in 1500 on average have laser treatment for short sight.\\
(i) Use a suitable approximation to find the probability that, of a random sample of 2700 contact lens wearers, more than 2 people have laser treatment.\\
(ii) In a random sample of $n$ contact lens wearers the probability that no one has laser treatment is less than 0.01 . Find the least possible value of $n$.
\hfill \mbox{\textit{CAIE S2 2004 Q5 [7]}}