| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.8 This is a multi-part Further Statistics question requiring (a) a two-tailed Poisson hypothesis test with critical region determination, (b) binomial probability calculation, and (c) reverse-engineering a Poisson parameter from a percentile condition. While each technique is standard for FS1, the combination of three distinct statistical methods and the non-trivial setup in part (c) requiring careful interpretation makes this moderately challenging, placing it above average difficulty. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(H_0 : \lambda = 1.7\) \(H_1 : \lambda \neq 1.7\) | B1 | 2.5 |
| \([X = \text{no. of calls in 10 mins}]\) \(X \sim \text{Po}(17)\) | M1 | 3.3 |
| \([\text{P}(X \geq 25) = 1 - \text{P}(X \leq 24)] = 0.0406463\ldots\) or CR: \(X \geq 27\) | A1 | 3.4 |
| \([0.04..> 0.025/ 25 \text{ is not in CR so not significant}]\) insufficient evidence of a change in rate of calls | A1 | 2.2b |
| (b) \([T = \text{no. of calls longer than 8 minutes}]\) \(T \sim \text{B}(70, 0.012)\) | M1 | 3.3 |
| \([\text{P}(T > 2) =] \text{P}(T \geq 3) = 1 - \text{P}(T \leq 2) = 1 - 0.947725\ldots = \text{awrt } \mathbf{0.0523}\) | M1, A1 | 3.4, 1.1b |
| (c) \([C = \text{no. of calls out of 900 longer than 30 mins}]\) \([C \sim \text{B}(900, p)]\) \(C \sim \text{Po}(900p)\) | M1 | 3.3 |
| \(\text{P}(C = 0) \approx e^{-900p} = 0.05\) \(900p = -\ln(0.05) [= 2.9957\ldots]\) \(p = 0.003328\ldots\) awrt \(\mathbf{0.00333}\) | M1, M1, A1 | 3.4, 1.1b, 1.1b |
**(a)** $H_0 : \lambda = 1.7$ $H_1 : \lambda \neq 1.7$ | B1 | 2.5 | B1 for both hypotheses correct which must be attached to $H_0$ and $H_1$ must be in terms of $\lambda$ or $\mu$ allow either 1.7 or 17
$[X = \text{no. of calls in 10 mins}]$ $X \sim \text{Po}(17)$ | M1 | 3.3 |
$[\text{P}(X \geq 25) = 1 - \text{P}(X \leq 24)] = 0.0406463\ldots$ or CR: $X \geq 27$ | A1 | 3.4 |
$[0.04..> 0.025/ 25 \text{ is not in CR so not significant}]$ insufficient evidence of a change in rate of calls | A1 | 2.2b | (4)
**(b)** $[T = \text{no. of calls longer than 8 minutes}]$ $T \sim \text{B}(70, 0.012)$ | M1 | 3.3 |
$[\text{P}(T > 2) =] \text{P}(T \geq 3) = 1 - \text{P}(T \leq 2) = 1 - 0.947725\ldots = \text{awrt } \mathbf{0.0523}$ | M1, A1 | 3.4, 1.1b | 1st M1 for sight or use of the correct binomial model. may be implied by sight of awrt 0.0406/7 or awrt 0.959 or 0.9747... or better; 1st A1 for correct prob of awrt 0.04; 2nd A1 (dep on M1A1) for a correct conclusion in context mentioning "rate of calls" o.e. Allow e.g. 'The rate of calls is 1.7 per minute/17 per 10 minutes' Must be rate o.e. not "number"; Δ0 if inconsistent comments are seen e.g. "reject $H_0$, no change in rate of calls" | (3)
**(c)** $[C = \text{no. of calls out of 900 longer than 30 mins}]$ $[C \sim \text{B}(900, p)]$ $C \sim \text{Po}(900p)$ | M1 | 3.3 |
$\text{P}(C = 0) \approx e^{-900p} = 0.05$ $900p = -\ln(0.05) [= 2.9957\ldots]$ $p = 0.003328\ldots$ awrt $\mathbf{0.00333}$ | M1, M1, A1 | 3.4, 1.1b, 1.1b | 1st M1 for sight or use of Po$(900p)$ (as a suitable approx. to B$(900, p)$) (may be implied by correct answer awrt 0.00333); 2nd M1 for a correct equation in $p$ or correct use of P$(C = 0)$ from Po e.g. $e^{-\lambda} = 0.05$; 3rd M1 for a correct method to solve for $p$ (allow $p = \pm \ln(0.05)/900$) or to solve for $\lambda$, i.e. $\lambda = \text{awrt } 3(00)$; A1 for $p = \text{awrt } 0.00333$ **Must see Po used** condone $\frac{1}{300}$ o.e. Allow standard form (awrt $3.33 \times 10^{-3}$) or percentage (awrt 0.333%) | (4)
**(Total: 11 marks)**
---\begin{enumerate}
\item Telephone calls arrive at a call centre randomly, at an average rate of 1.7 per minute. After the call centre was closed for a week, in a random sample of 10 minutes there were 25 calls to the call centre.\\
(a) Carry out a suitable test to determine whether or not there is evidence that the rate of calls arriving at the call centre has changed.\\
Use a $5 \%$ level of significance and state your hypotheses clearly.
\end{enumerate}
Only 1.2\% of the calls to the call centre last longer than 8 minutes.\\
One day Tiang has 70 calls.\\
(b) Find the probability that out of these 70 calls Tiang has more than 2 calls lasting longer than 8 minutes.
The call centre records show that $95 \%$ of days have at least one call lasting longer than 30 minutes.\\
On Wednesday 900 calls arrived at the call centre and none of them lasted longer than 30 minutes.\\
(c) Use a Poisson approximation to estimate the proportion of calls arriving at the call centre that last longer than 30 minutes.
\hfill \mbox{\textit{Edexcel FS1 2023 Q2 [11]}}