| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.8 This question requires understanding of Poisson scaling across time periods, hypothesis testing framework including Type I and Type II errors, and conceptual understanding of when errors can occur. Part (i) is routine calculation, but parts (ii)-(iv) require proper setup of hypotheses, critical region identification, and deeper conceptual reasoning about error types—significantly above average for A-level statistics. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda = 4.65\) | B1 | |
| \(e^{-4.65} \times \frac{4.65^4}{4!}\) | M1 | Poisson \(P(X=4)\) with any \(\lambda\) |
| \(= 0.186\) (3 sf) | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda = 3.875\) | B1 | |
| \(= e^{-3.875}\left(1 + 3.875 + \frac{3.875^2}{2!}\right) = 0.257\) (3 sf) | M1, A1 | \(P(X = 0, 1, 2)\); Attempted, any \(\lambda\); As final answer. Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda = 1.5\) | B1 | |
| \(1 - e^{-1.5}\left(1 + 1.5 + \frac{1.5^2}{2!}\right)\) | M1 | \(1 - P(X = 0, 1, 2)\); Attempted, any \(\lambda\) |
| \(= 0.191\) (3 sf) | A1 | As final answer. Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| He will reject \(H_0\). | B1 | Total: 1 |
## Question 6(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = 4.65$ | B1 | |
| $e^{-4.65} \times \frac{4.65^4}{4!}$ | M1 | Poisson $P(X=4)$ with any $\lambda$ |
| $= 0.186$ (3 sf) | A1 | **Total: 3** |
## Question 6(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = 3.875$ | B1 | |
| $= e^{-3.875}\left(1 + 3.875 + \frac{3.875^2}{2!}\right) = 0.257$ (3 sf) | M1, A1 | $P(X = 0, 1, 2)$; Attempted, any $\lambda$; As final answer. **Total: 3** |
## Question 6(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = 1.5$ | B1 | |
| $1 - e^{-1.5}\left(1 + 1.5 + \frac{1.5^2}{2!}\right)$ | M1 | $1 - P(X = 0, 1, 2)$; Attempted, any $\lambda$ |
| $= 0.191$ (3 sf) | A1 | As final answer. **Total: 3** |
## Question 6(iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| He will reject $H_0$. | B1 | **Total: 1** |6 The number of accidents on a certain road has a Poisson distribution with mean 3.1 per 12-week period.\\
(i) Find the probability that there will be exactly 4 accidents during an 18-week period.
Following the building of a new junction on this road, an officer wishes to determine whether the number of accidents per week has decreased. He chooses 15 weeks at random and notes the number of accidents. If there are fewer than 3 accidents altogether he will conclude that the number of accidents per week has decreased. He assumes that a Poisson distribution still applies.\\
(ii) Find the probability of a Type I error.\\
(iii) Given that the mean number of accidents per week is now 0.1 , find the probability of a Type II error.\\
(iv) Given that there were 2 accidents during the 15 weeks, explain why it is impossible for the officer to make a Type II error.
\hfill \mbox{\textit{CAIE S2 2014 Q6 [10]}}