| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | State Poisson approximation with justification |
| Difficulty | Moderate -0.5 This is a straightforward application of the Poisson approximation to the binomial distribution. Part (i) requires stating the standard conditions (n large, p small, np moderate) and a simple probability calculation. Part (ii) involves combining two Poisson distributions, which is routine knowledge at this level. The question requires recall of when the approximation is valid and basic Poisson probability calculations, but no problem-solving insight or complex multi-step reasoning. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (Bin) with \(n > 50\) and mean (or \(np\)) \(< 5\) | B1 | Accept n 'large', p 'small' |
| \(Po(1.5)\) | B1 | Poisson with correct mean stated or implied |
| \(1 - e^{-1.5}\) | M1 | Poisson \(1 - P(X=0)\); allow incorrect \(\lambda\); allow 1 end error |
| \(= 0.777\) (3 sf) | A1 | SR If zero scored use of Bin leading to 0.778/0.779 scores B1. Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3.5\) | B1 | Correct mean stated or implied |
| \(e^{-3.5}\left(\frac{3.5^4}{4!} + \frac{3.5^5}{5!} + \frac{3.5^6}{6!}\right)\) | M1 | Poisson \(P(X = 4, 5, 6)\); allow incorrect \(\lambda\); allow 1 end error |
| \(= 0.398\) (3 sf) | A1 | Total: 3 |
## Question 2(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| (Bin) with $n > 50$ and mean (or $np$) $< 5$ | B1 | Accept n 'large', p 'small' |
| $Po(1.5)$ | B1 | Poisson with correct mean stated or implied |
| $1 - e^{-1.5}$ | M1 | Poisson $1 - P(X=0)$; allow incorrect $\lambda$; allow 1 end error |
| $= 0.777$ (3 sf) | A1 | SR If zero scored use of Bin leading to 0.778/0.779 scores B1. **Total: 4** |
## Question 2(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3.5$ | B1 | Correct mean stated or implied |
| $e^{-3.5}\left(\frac{3.5^4}{4!} + \frac{3.5^5}{5!} + \frac{3.5^6}{6!}\right)$ | M1 | Poisson $P(X = 4, 5, 6)$; allow incorrect $\lambda$; allow 1 end error |
| $= 0.398$ (3 sf) | A1 | **Total: 3** |
---2 The probability that a randomly chosen plant of a certain kind has a particular defect is 0.01 . A random sample of 150 plants is taken.\\
(i) Use an appropriate approximating distribution to find the probability that at least 1 plant has the defect. Justify your approximating distribution.
The probability that a randomly chosen plant of another kind has the defect is 0.02 . A random sample of 100 of these plants is taken.\\
(ii) Use an appropriate approximating distribution to find the probability that the total number of plants with the defect in the two samples together is more than 3 and less than 7 .
\hfill \mbox{\textit{CAIE S2 2014 Q2 [7]}}