| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Finding minimum stock level for demand |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with standard probability calculations. Part (i) is direct formula substitution, part (ii) requires P(X>4)=1-P(X≤4) using tables, and part (iii) involves systematic trial with cumulative probabilities until finding the threshold—routine for S2 level with no conceptual challenges beyond following the method. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(2) = e^{-2.2} \times \frac{2.2^2}{2} = 0.209\) | M1, A1 (2 marks) | For a Poisson attempt; For correct answer |
| (ii) \(P(X > 4) = 1 - P(X = 0,1,2,3,4) = 1 - e^{-2.2}(1 + 3.2 + 3.2^2/2 + 3.2^3/6 + 3.2^4/24)\) | M1, M1, A1, A1 (4 marks) | For realising that \(P(X > 4)\) is required; For an attempt to evaluate this probability as \(1 - ...\); For correct unsimplified expression; Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) by trial and error \(P(X > 5) = 0.105\), \(P(X > 6) = 0.0446\) which is < 5%, \(n = 6\) | M1, A1, M1, A1 (4 marks) | For any sensible attempt; For finding correct \(P(X > 5)\); For finding correct \(P(X > 6)\); Correct answer |
**(i)** $P(2) = e^{-2.2} \times \frac{2.2^2}{2} = 0.209$ | M1, A1 (2 marks) | For a Poisson attempt; For correct answer
**(ii)** $P(X > 4) = 1 - P(X = 0,1,2,3,4) = 1 - e^{-2.2}(1 + 3.2 + 3.2^2/2 + 3.2^3/6 + 3.2^4/24)$ | M1, M1, A1, A1 (4 marks) | For realising that $P(X > 4)$ is required; For an attempt to evaluate this probability as $1 - ...$; For correct unsimplified expression; Correct answer
$= 0.219$
**(iii)** by trial and error $P(X > 5) = 0.105$, $P(X > 6) = 0.0446$ which is < 5%, $n = 6$ | M1, A1, M1, A1 (4 marks) | For any sensible attempt; For finding correct $P(X > 5)$; For finding correct $P(X > 6)$; Correct answer
---6 A shopkeeper sells electric fans. The demand for fans follows a Poisson distribution with mean 3.2 per week.\\
(i) Find the probability that the demand is exactly 2 fans in any one week.\\
(ii) The shopkeeper has 4 fans in his shop at the beginning of a week. Find the probability that this will not be enough to satisfy the demand for fans in that week.\\
(iii) Given instead that he has $n$ fans in his shop at the beginning of a week, find, by trial and error, the least value of $n$ for which the probability of his not being able to satisfy the demand for fans in that week is less than 0.05 .
\hfill \mbox{\textit{CAIE S2 2005 Q6 [10]}}