| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Multiple periods with binomial structure |
| Difficulty | Standard +0.3 This is a straightforward multi-part Poisson question requiring: (a) direct use of Poisson tables/calculator for standard probability calculations, (b) normal approximation to Poisson (standard S2 technique), and (c) binomial calculation using the result from (b). All steps are routine applications of standard methods with no novel insight required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X\) represents number of telephone calls per min \(\Rightarrow X \sim Po(9)\) | ||
| \(P(X > 5) = 1 - P(X \leq 5)\) | M1 | For using or writing \(1 - P(X \leq 5)\) or \(1 - P(X < 6)\), may be implied by awrt 0.884 |
| \(= 0.8843\) | A1 | awrt 0.884 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(4 \leq X < 10) = P(X \leq 9) - P(X \leq 3)\) | M1 | For using or writing \(P(X \leq 9) - P(X \leq 3)\), oe may be implied by awrt 0.566 |
| \(= 0.5874 - 0.0212 = 0.5662\) | A1 | awrt 0.566 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(D\) represents number of telephone calls per day | ||
| Normal approximation \(\mu = \frac{7 \times 60 \times 9}{10} = 378\) and \(\sigma^2 = 378\) | M1 | Using normal approximation with mean = variance = 378 or sd = \(\sqrt{378}\) (awrt 19.4), or writing N(378,378) |
| \(P(D < 370) \approx P\left(Z < \frac{369.5 - 378}{\sqrt{378}}\right)\) | M1, M1d, A1ft | standardise \(\pm 0.5\); M1d dep on previous M mark; A1ft standardisation with correct CC ie \(\pm\frac{369.5 - \text{"their378"}}{\sqrt{\text{"their378"}}}\) or awrt \(\pm 0.44\) |
| \(\approx P(Z < -0.44)\) | ||
| \(= 1 - 0.670\) | ||
| \(= 0.33\) or \(0.330\) or awrt \(0.331\) | A1 | 0.33 must be from correct standardisation. 0.33 with no working gains NO marks. 0.330 or 0.331 with no working gains full marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(W\) represents number of days which have fewer than 370 telephone calls | ||
| \(W \sim B(5, \text{"0.33"})\) or \(W \sim B(5, \text{"0.67"})\) | M1 | Writing \(B(5, \text{"0.33"})\) or \(B(5, 1-\text{"0.33"})\) or seeing \({}^5C_n(\text{"0.33"})^n(1-\text{"0.33"})^{5-n}\) where \(1 \leq n \leq 4\) |
| \(P(W=4) + P(W=5)\) or \(P(W=0) + P(W=1)\) | M1 | \(1-(1-\text{"0.33"})^5 - 5(\text{"0.33"})^1(1-\text{"0.33"})^4 - 10(\text{"0.33"})^2(1-\text{"0.33"})^3 - 10(\text{"0.33"})^3(1-\text{"0.33"})^2\) |
| \(= 5(\text{"0.33"})^4(1-\text{"0.33"}) + (\text{"0.33"})^5\) or \((1-\text{"0.67"})^5 + 5(\text{"0.67"})("1-0.67")^4\) | oe Allow if using \({}^nC_r\) form or factorial form | |
| \(= 0.0436\) | A1 | awrt 0.044; awrt 0.044 with no incorrect working gains M1M1A1 |
# Question 1:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X$ represents number of telephone calls per min $\Rightarrow X \sim Po(9)$ | | |
| $P(X > 5) = 1 - P(X \leq 5)$ | M1 | For using or writing $1 - P(X \leq 5)$ or $1 - P(X < 6)$, may be implied by awrt 0.884 |
| $= 0.8843$ | A1 | awrt 0.884 |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(4 \leq X < 10) = P(X \leq 9) - P(X \leq 3)$ | M1 | For using or writing $P(X \leq 9) - P(X \leq 3)$, oe may be implied by awrt 0.566 |
| $= 0.5874 - 0.0212 = 0.5662$ | A1 | awrt 0.566 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $D$ represents number of telephone calls per day | | |
| Normal approximation $\mu = \frac{7 \times 60 \times 9}{10} = 378$ and $\sigma^2 = 378$ | M1 | Using normal approximation with mean = variance = 378 or sd = $\sqrt{378}$ (awrt 19.4), or writing N(378,378) |
| $P(D < 370) \approx P\left(Z < \frac{369.5 - 378}{\sqrt{378}}\right)$ | M1, M1d, A1ft | standardise $\pm 0.5$; M1d dep on previous M mark; A1ft standardisation with correct CC ie $\pm\frac{369.5 - \text{"their378"}}{\sqrt{\text{"their378"}}}$ or awrt $\pm 0.44$ |
| $\approx P(Z < -0.44)$ | | |
| $= 1 - 0.670$ | | |
| $= 0.33$ or $0.330$ or awrt $0.331$ | A1 | **0.33 must be from correct standardisation. 0.33 with no working gains NO marks. 0.330 or 0.331 with no working gains full marks.** |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $W$ represents number of days which have fewer than 370 telephone calls | | |
| $W \sim B(5, \text{"0.33"})$ or $W \sim B(5, \text{"0.67"})$ | M1 | Writing $B(5, \text{"0.33"})$ or $B(5, 1-\text{"0.33"})$ or seeing ${}^5C_n(\text{"0.33"})^n(1-\text{"0.33"})^{5-n}$ where $1 \leq n \leq 4$ |
| $P(W=4) + P(W=5)$ or $P(W=0) + P(W=1)$ | M1 | $1-(1-\text{"0.33"})^5 - 5(\text{"0.33"})^1(1-\text{"0.33"})^4 - 10(\text{"0.33"})^2(1-\text{"0.33"})^3 - 10(\text{"0.33"})^3(1-\text{"0.33"})^2$ |
| $= 5(\text{"0.33"})^4(1-\text{"0.33"}) + (\text{"0.33"})^5$ or $(1-\text{"0.67"})^5 + 5(\text{"0.67"})("1-0.67")^4$ | | oe Allow if using ${}^nC_r$ form or factorial form |
| $= 0.0436$ | A1 | awrt 0.044; **awrt 0.044 with no incorrect working gains M1M1A1** |
---\begin{enumerate}
\item In a call centre, the number of telephone calls, $X$, received during any 10 -minute period follows a Poisson distribution with mean 9\\
(a) Find\\
(i) $\mathrm { P } ( X > 5 )$\\
(ii) $\mathrm { P } ( 4 \leqslant X < 10 )$
\end{enumerate}
The length of a working day is 7 hours.\\
(b) Using a suitable approximation, find the probability that there are fewer than 370 telephone calls in a randomly selected working day.
A week, consisting of 5 working days, is selected at random.\\
(c) Find the probability that in this week at least 4 working days have fewer than 370 telephone calls.
\hfill \mbox{\textit{Edexcel S2 2018 Q1 [12]}}