| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Moderate -0.8 This is a straightforward Poisson distribution question requiring only standard recall and application. Part (i) asks for textbook conditions (random, independent events at constant rate), part (ii) is direct substitution into the Poisson formula with λ=10, and part (iii) uses the standard normal approximation technique taught in S2. No problem-solving insight or complex multi-step reasoning required. |
| 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 |
|---|---|---|
| (i) people call randomly, independently, at an average uniform rate | B1, B1 | Any two seen, or equivalent words in context. SR If B0 B0 scored and two correct reasons seen but not in context score B1. |
| (ii) \(P(8) = \frac{e^{-10}10^8}{8!} = 0.113\) | M1, A1 | 10 seen in a Poisson calculation. Correct answer. |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=250) = \Phi\left(\frac{250.5-240}{\sqrt{240}}\right) - \Phi\left(\frac{249.5-240}{\sqrt{240}}\right) = \Phi(0.678) - \Phi(0.613) = 0.7512 - 0.7301 = 0.0211\) (accept 3sf or more rounding to 0.021) | B1, M1, M1, M1, A1 | mean and variance = 240. Standardising with their mean and variance. Subtracting the two relevant Φs. Correct answer. SR If 0/4 scored P(250)=\(e^{-240}.240^{250}/250!\) seen scores B1. |
**(i)** people call randomly, independently, at an average uniform rate | B1, B1 | Any two seen, or equivalent words in context. SR If B0 B0 scored and two correct reasons seen but not in context score B1.
**(ii)** $P(8) = \frac{e^{-10}10^8}{8!} = 0.113$ | M1, A1 | 10 seen in a Poisson calculation. Correct answer.
**(iii)** $X \sim N(240, 240)$
$P(X=250) = \Phi\left(\frac{250.5-240}{\sqrt{240}}\right) - \Phi\left(\frac{249.5-240}{\sqrt{240}}\right) = \Phi(0.678) - \Phi(0.613) = 0.7512 - 0.7301 = 0.0211$ (accept 3sf or more rounding to 0.021) | B1, M1, M1, M1, A1 | mean and variance = 240. Standardising with their mean and variance. Subtracting the two relevant Φs. Correct answer. SR If 0/4 scored P(250)=$e^{-240}.240^{250}/250!$ seen scores B1.
---5 It is proposed to model the number of people per hour calling a car breakdown service between the times 0900 and 2100 by a Poisson distribution.\\
(i) Explain why a Poisson distribution may be appropriate for this situation.
People call the car breakdown service at an average rate of 20 per hour, and a Poisson distribution may be assumed to be a suitable model.\\
(ii) Find the probability that exactly 8 people call in any half hour.\\
(iii) By using a suitable approximation, find the probability that exactly 250 people call in the 12 hours between 0900 and 2100.
\hfill \mbox{\textit{CAIE S2 2007 Q5 [8]}}