| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.3 This is a straightforward hypothesis test using the Poisson distribution with clearly defined parameters. Part (a) requires standard one-tailed test procedure (stating hypotheses, calculating probability, comparing to significance level), while part (b) involves finding a critical value by trial. Both parts are routine S2 applications requiring no novel insight, though slightly easier than average due to the clear setup and standard methodology. |
| Spec | 2.05b Hypothesis test for binomial proportion5.02i Poisson distribution: random events model |
| Answer | Marks |
|---|---|
| (a) \(X \sim \text{Po}(\lambda)\) | B1 B1 |
| \(H_0: \lambda = 3.5\), \(H_1: \lambda > 3.5\) | M1 A1 A1 |
| Under \(H_0\), \(P(X \geq 7) = 1 - 0.9347 > 5\%\), so do not reject \(H_0\) | M1 A1 |
| (b) Need \(P(X \geq n) < 0.001\) and \(P(X \leq n-1) > 0.999\) | M1 A1 |
| \(n - 1 = 11\), so 12 calls are required | A1 |
| 8 |
(a) $X \sim \text{Po}(\lambda)$ | B1 B1 |
$H_0: \lambda = 3.5$, $H_1: \lambda > 3.5$ | M1 A1 A1 |
Under $H_0$, $P(X \geq 7) = 1 - 0.9347 > 5\%$, so do not reject $H_0$ | M1 A1 |
(b) Need $P(X \geq n) < 0.001$ and $P(X \leq n-1) > 0.999$ | M1 A1 |
$n - 1 = 11$, so 12 calls are required | A1 |
| | 8 |An insurance company conducts its business by using a Call Centre. The average number of calls per minute is 3.5. In the first minute after a TV advertisement is shown, the number of calls received is 7.
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses carefully, and working at the 5\% significance level, test whether the advertisement has had an effect. [5 marks]
\item Find the number of calls that would be required in the first minute for the null hypothesis to be rejected at the 0.1\% significance level. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q2 [8]}}