| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Sum of three or more Poissons |
| Difficulty | Standard +0.3 This question tests standard Poisson distribution properties: basic probability calculations, sum of independent Poisson variables, and solving an inequality involving Poisson probabilities. All parts are routine applications of formulas with no novel insight required. Part (iii) involves logarithms but is still straightforward. Slightly above average due to multiple parts and the need to combine Poisson distributions, but well within typical S2 scope. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 - e^{-0.8}(1 + 0.8) = 0.191\) (3 sf) | M1, A1 (2) | Allow one end error |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda = 3 \times 0.8 + 2 \times 2.7 = 7.8\) | M1 | Attempt find \(\lambda\) |
| \(e^{-7.8}\left(1 + 7.8 + \frac{7.8^2}{2} + \frac{7.8^3}{3!} + \frac{7.8^4}{4!}\right)\) | M1 | \(P(0,1,2,3,4)\) using their \(\lambda\), allow one end error |
| \(= 0.112\) (3 sf) | A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e^{-0.8n} < 0.1\), allow '\(=\)' | M1* | or \(e^{-x} < 0.1\) — M1 for trial and improvement to find \(P(0)\): \(e^{-1.6} = 0.20\), \(n=2\) |
| \(-0.8n < \ln 0.1\), allow '\(=\)' | M1* dep | \(-x < \ln 0.1\) — \(e^{-2.4} = 0.09\), \(n=3\); M1* dep try |
| \(\min n = 3\) | A1 (3) | Correctly obtained; \(n = 3\) both correct — A1 |
## Question 4:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 - e^{-0.8}(1 + 0.8) = 0.191$ (3 sf) | M1, A1 (2) | Allow one end error |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = 3 \times 0.8 + 2 \times 2.7 = 7.8$ | M1 | Attempt find $\lambda$ |
| $e^{-7.8}\left(1 + 7.8 + \frac{7.8^2}{2} + \frac{7.8^3}{3!} + \frac{7.8^4}{4!}\right)$ | M1 | $P(0,1,2,3,4)$ using their $\lambda$, allow one end error |
| $= 0.112$ (3 sf) | A1 (3) | |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e^{-0.8n} < 0.1$, allow '$=$' | M1* | or $e^{-x} < 0.1$ — M1 for trial and improvement to find $P(0)$: $e^{-1.6} = 0.20$, $n=2$ |
| $-0.8n < \ln 0.1$, allow '$=$' | M1* dep | $-x < \ln 0.1$ — $e^{-2.4} = 0.09$, $n=3$; M1* dep try |
| $\min n = 3$ | A1 (3) | Correctly obtained; $n = 3$ both correct — A1 |
---4 The number of lions seen per day during a standard safari has the distribution $\operatorname { Po } ( 0.8 )$. The number of lions seen per day during an off-road safari has the distribution $\operatorname { Po } ( 2.7 )$. The two distributions are independent.\\
(i) Susan goes on a standard safari for one day. Find the probability that she sees at least 2 lions.\\
(ii) Deena goes on a standard safari for 3 days and then on an off-road safari for 2 days. Find the probability that she sees a total of fewer than 5 lions.\\
(iii) Khaled goes on a standard safari for $n$ days, where $n$ is an integer. He wants to ensure that his chance of not seeing any lions is less than $10 \%$. Find the smallest possible value of $n$.
\hfill \mbox{\textit{CAIE S2 2012 Q4 [8]}}