| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Basic sum of two Poissons |
| Difficulty | Standard +0.3 This is a straightforward Poisson distribution question requiring rate scaling (8 per year to different time periods) and standard probability calculations. Part (i) is routine P(X>3), part (ii) requires recognizing that the sum of independent Poisson variables is Poisson, and part (iii) uses normal approximation. All techniques are standard S2 material with no novel insight required, making it slightly easier than average. |
| 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) \(\lambda = 2\), \(P(X > 3) = 1 - P(0, 1, 2, 3)\) | B1 | Correct mean (used) |
| \(= 1 - e^{-2}\left(1 + 2 + \frac{2^2}{2} + \frac{2^3}{3!}\right) = 1 - 0.857 = 0.143\) | M1, A1 | Poisson \(1 - P(0,1,2,3)\) or \(P(0,1,2)\) or \(P(1,2,3)\); Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(\lambda = 16/3\), \(P(7) = e^{-16/3}\left(\frac{(16/3)^7}{7!}\right) = 0.118\) | B1, M1, A1 | Correct new mean; P(7) using a different mean from (i); Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) \(X \sim N(160, 160)\) | B1, M1 | Correct mean and variance; Standardising attempt with or without cc must have sq rt |
| \(P(X < 137) = P\left(z < \frac{136.5 - 160}{\sqrt{160}}\right) = P(z < -1.858) = 1 - 0.9684 = 0.0316\) | M1, A1 | Cc of 136.5 or 137.5 and area < 0.5; Correct answer |
(i) $\lambda = 2$, $P(X > 3) = 1 - P(0, 1, 2, 3)$ | B1 | Correct mean (used)
$= 1 - e^{-2}\left(1 + 2 + \frac{2^2}{2} + \frac{2^3}{3!}\right) = 1 - 0.857 = 0.143$ | M1, A1 | Poisson $1 - P(0,1,2,3)$ or $P(0,1,2)$ or $P(1,2,3)$; Correct answer
**[3]**
(ii) $\lambda = 16/3$, $P(7) = e^{-16/3}\left(\frac{(16/3)^7}{7!}\right) = 0.118$ | B1, M1, A1 | Correct new mean; P(7) using a different mean from (i); Correct final answer
**[3]**
(iii) $X \sim N(160, 160)$ | B1, M1 | Correct mean and variance; Standardising attempt with or without cc must have sq rt
$P(X < 137) = P\left(z < \frac{136.5 - 160}{\sqrt{160}}\right) = P(z < -1.858) = 1 - 0.9684 = 0.0316$ | M1, A1 | Cc of 136.5 or 137.5 and area < 0.5; Correct answer
**[4]**3 Major avalanches can be regarded as randomly occurring events. They occur at a uniform average rate of 8 per year.\\
(i) Find the probability that more than 3 major avalanches occur in a 3-month period.\\
(ii) Find the probability that any two separate 4 -month periods have a total of 7 major avalanches.\\
(iii) Find the probability that a total of fewer than 137 major avalanches occur in a 20 -year period.
\hfill \mbox{\textit{CAIE S2 2009 Q3 [10]}}