| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Finding maximum n for P(X=0) threshold |
| Difficulty | Standard +0.3 This is a standard S2 Poisson distribution question with routine calculations: part (a) requires direct use of Poisson probability formulas with adjusted rates, part (b) involves iterative solving which is slightly more demanding, and part (c) is a straightforward hypothesis test. All techniques are textbook exercises with no novel insight required, making it slightly easier than average overall. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X \geq 4) = 1 - P(X \leq 3)\) | M1 | Writing or using \(1 - P(X \leq 3)\); implied by awrt 0.242 |
| \(= 1 - 0.7576\) | ||
| \(= 0.2424\) | A1 | awrt 0.242 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using \(Po(0.625)\) | B1 | |
| \(P(X=3) = \frac{e^{-0.625}0.625^3}{3!}\) | M1 | Finding \(P(X=3)\) with any \(\lambda\), e.g. \(\frac{e^{-\lambda}\lambda^3}{3!}\) or \(P(X \leq 3) - P(X \leq 2)\); implied by awrt 0.0218 |
| \(= 0.02177...\) | A1 | awrt 0.0218 (5 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1 - P(X=0) < 0.2\) or \(P(X=0) > 0.8\) | M1 | Writing or using \(1 - P(X=0) < 0.2\) or \(P(X=0) > 0.8\); allow use of \(=\) instead of \(>\) or \(<\); implied by \(e^{-l} = 0.8\) or \(e^{-l} > 0.8\) or awrt 5.36 or 0.089 |
| \(e^{-2.5t} > 0.8\) | M1 | Writing an inequality of the form \(e^{-l} > 0.8\) using any \(l\); may be implied by awrt 5.36 or 0.089; do not allow \(e^{-l} = 0.8\) |
| \(t < 0.089\ldots\) hours \(= 5.36\) mins | ||
| \([t <]\ 5\) mins | A1cso | Both method marks must be awarded; accept 5 or \(t = 5\) or \(t < 5\) (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda = 2.5\ (\lambda = 5)\); \(H_1: \lambda > 2.5\ (\lambda > 5)\) | B1 | Both hypotheses using \(\lambda\) or \(\mu\); allow 5 or 2.5; must be clear which is \(H_0\) and which is \(H_1\) |
| \(P(X \geq 10) = 1 - P(X \leq 9)\) | M1 | Writing or using \(Po(5)\) and \(1 - P(X \leq 9)\); may be implied by a correct CR; do not allow writing \(P(X \geq 10)\) |
| \(= 1 - 0.9682\) | ||
| \(= 0.0318\) | A1 | awrt 0.0318; allow CR \(X \geq 10\) or \(X > 9\) |
| NB allow M1A1 if not using CR route for \(P(X \leq 9) =\) awrt 0.968 | ||
| Sufficient evidence to reject \(H_0\), accept \(H_1\); significant; 10 does lie in the critical region | M1d | Dependent on previous M; correct statement (do not allow if contradicting non-contextual statements); ft their Prob/CR compared with 0.05/10 (0.95 if using 0.968) |
| There is sufficient evidence that the mean rate of telephone calls has increased | A1cso | Must include word calls and idea that rate has increased; do not allow "it has changed" on its own; all previous marks must be awarded; M1A1 awarded for correct contextual statement on its own provided previous marks have been awarded (5 marks) |
## Question 2:
### Part (a)(i)
$X \sim Po(2.5)$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X \geq 4) = 1 - P(X \leq 3)$ | M1 | Writing or using $1 - P(X \leq 3)$; implied by awrt 0.242 |
| $= 1 - 0.7576$ | | |
| $= 0.2424$ | A1 | awrt 0.242 |
### Part (a)(ii)
$X \sim Po(0.625)$
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $Po(0.625)$ | B1 | |
| $P(X=3) = \frac{e^{-0.625}0.625^3}{3!}$ | M1 | Finding $P(X=3)$ with any $\lambda$, e.g. $\frac{e^{-\lambda}\lambda^3}{3!}$ or $P(X \leq 3) - P(X \leq 2)$; implied by awrt 0.0218 |
| $= 0.02177...$ | A1 | awrt 0.0218 **(5 marks total)** |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 - P(X=0) < 0.2$ or $P(X=0) > 0.8$ | M1 | Writing or using $1 - P(X=0) < 0.2$ or $P(X=0) > 0.8$; allow use of $=$ instead of $>$ or $<$; implied by $e^{-l} = 0.8$ or $e^{-l} > 0.8$ or awrt 5.36 or 0.089 |
| $e^{-2.5t} > 0.8$ | M1 | Writing an inequality of the form $e^{-l} > 0.8$ using any $l$; may be implied by awrt 5.36 or 0.089; do not allow $e^{-l} = 0.8$ |
| $t < 0.089\ldots$ hours $= 5.36$ mins | | |
| $[t <]\ 5$ mins | A1cso | Both method marks must be awarded; accept 5 or $t = 5$ or $t < 5$ **(3 marks)** |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 2.5\ (\lambda = 5)$; $H_1: \lambda > 2.5\ (\lambda > 5)$ | B1 | Both hypotheses using $\lambda$ or $\mu$; allow 5 or 2.5; must be clear which is $H_0$ and which is $H_1$ |
| $P(X \geq 10) = 1 - P(X \leq 9)$ | M1 | Writing or using $Po(5)$ and $1 - P(X \leq 9)$; may be implied by a correct CR; do not allow writing $P(X \geq 10)$ |
| $= 1 - 0.9682$ | | |
| $= 0.0318$ | A1 | awrt 0.0318; allow CR $X \geq 10$ or $X > 9$ |
| | | **NB** allow M1A1 if not using CR route for $P(X \leq 9) =$ awrt 0.968 |
| Sufficient evidence to reject $H_0$, accept $H_1$; significant; 10 does lie in the critical region | M1d | Dependent on previous M; correct statement (do not allow if contradicting non-contextual statements); ft their Prob/CR compared with 0.05/10 (0.95 if using 0.968) |
| There is sufficient evidence that the mean rate of telephone **calls** has increased | A1cso | Must include word **calls** and idea that rate has increased; do not allow "it has changed" on its own; all previous marks must be awarded; M1A1 awarded for correct contextual statement on its own provided previous marks have been awarded **(5 marks)** |
---\begin{enumerate}
\item A company receives telephone calls at random at a mean rate of 2.5 per hour.\\
(a) Find the probability that the company receives\\
(i) at least 4 telephone calls in the next hour,\\
(ii) exactly 3 telephone calls in the next 15 minutes.\\
(b) Find, to the nearest minute, the maximum length of time the telephone can be left unattended so that the probability of missing a telephone call is less than 0.2
\end{enumerate}
The company puts an advert in the local newspaper. The number of telephone calls received in a randomly selected 2 hour period after the paper is published is 10\\
(c) Test at the 5\% level of significance whether or not the mean rate of telephone calls has increased. State your hypotheses clearly.\\
\hfill \mbox{\textit{Edexcel S2 2017 Q2 [13]}}