| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Joint probability of independent events |
| Difficulty | Standard +0.3 This is a straightforward application of the Poisson distribution requiring students to adjust the rate parameter for different time periods and calculate basic probabilities. Part (a) involves scaling the rate and using P(X > 4), part (b) requires binomial probability with Poisson probabilities, and part (c) needs independence of Poisson processes. While multi-part, each step follows standard procedures with no novel insight required, making it slightly easier than average for Further Maths. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(C\) = no. of calls in a 20 min period, \(C \sim Po(\ldots)\) | M1 | 1st M1 for selecting a Poisson model – written or used. May be implied by 2nd M1 or a correct answer |
| 80 calls per 4-hour period gives \(\frac{20}{3}\) per 20 mins, i.e. \(C \sim Po\!\left(\frac{20}{3}\right)\) | M1 | 2nd M1 for the correct Poisson \(Po\!\left(\frac{20}{3}\right)\) or \(Po(6.67)\) or better seen, and writing or using \(1 - P(C \leqslant 4)\) |
| \(P(C > 4) = 1 - P(C \leqslant 4) = 0.79437\ldots\) awrt \(\mathbf{0.794}\) | A1 | For awrt \(0.794\) (correct answer with no incorrect working scores 3/3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X\) = no. of 5 min periods with no calls, \(X \sim B(4, e^{-\frac{5}{3}})\) | M1 | For selecting a correct model \(B(4, 0.189)\) or better (calc: \(0.188875\ldots\)) |
| \(P(X=3) = 0.02186125\ldots\) awrt \(\mathbf{0.0219}\) | A1 | For using the model to get awrt \(0.0219\) (correct answer with no incorrect working scores 2/2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{exactly one call}) = e^{-\frac{5}{3}} \times \frac{5}{3}\) or \(e^{-5} \times 5\) | M1 | 1st M1 for a correct prob of 1 call (expressions in \(e\) or values). Allow \(0.31479\ldots\) or awrt \(0.315\) or \(0.033689\ldots\) or awrt \(0.0337\) |
| \(P(\text{exactly one call in each break}) = \left(e^{-\frac{5}{3}} \times \frac{5}{3}\right) \times \left(e^{-5} \times 5\right)\) | M1 | 2nd M1 for a correct probability statement or expression |
| \(= 0.0106052\ldots\) awrt \(\mathbf{0.0106}\) | A1 | For awrt \(0.0106\) (correct answer with no incorrect working scores 3/3) |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $C$ = no. of calls in a 20 min period, $C \sim Po(\ldots)$ | M1 | 1st M1 for selecting a Poisson model – written or used. May be implied by 2nd M1 or a correct answer |
| 80 calls per 4-hour period gives $\frac{20}{3}$ per 20 mins, i.e. $C \sim Po\!\left(\frac{20}{3}\right)$ | M1 | 2nd M1 for the correct Poisson $Po\!\left(\frac{20}{3}\right)$ or $Po(6.67)$ or better seen, and writing or using $1 - P(C \leqslant 4)$ |
| $P(C > 4) = 1 - P(C \leqslant 4) = 0.79437\ldots$ awrt $\mathbf{0.794}$ | A1 | For awrt $0.794$ (correct answer with no incorrect working scores 3/3) |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X$ = no. of 5 min periods with no calls, $X \sim B(4, e^{-\frac{5}{3}})$ | M1 | For selecting a correct model $B(4, 0.189)$ or better (calc: $0.188875\ldots$) |
| $P(X=3) = 0.02186125\ldots$ awrt $\mathbf{0.0219}$ | A1 | For using the model to get awrt $0.0219$ (correct answer with no incorrect working scores 2/2) |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{exactly one call}) = e^{-\frac{5}{3}} \times \frac{5}{3}$ or $e^{-5} \times 5$ | M1 | 1st M1 for a correct prob of 1 call (expressions in $e$ or values). Allow $0.31479\ldots$ or awrt $0.315$ or $0.033689\ldots$ or awrt $0.0337$ |
| $P(\text{exactly one call in each break}) = \left(e^{-\frac{5}{3}} \times \frac{5}{3}\right) \times \left(e^{-5} \times 5\right)$ | M1 | 2nd M1 for a correct probability statement or expression |
| $= 0.0106052\ldots$ awrt $\mathbf{0.0106}$ | A1 | For awrt $0.0106$ (correct answer with no incorrect working scores 3/3) |
---\begin{enumerate}
\item Indre works on reception in an office and deals with all the telephone calls that arrive. Calls arrive randomly and, in a 4-hour morning shift, there are on average 80 calls.\\
(a) Using a suitable model, find the probability of more than 4 calls arriving in a particular 20 -minute period one morning.
\end{enumerate}
Indre is allowed 20 minutes of break time during each 4-hour morning shift, which she can take in 5 -minute periods. When she takes a break, a machine records details of any call in the office that Indre has missed.
One morning Indre took her break time in 4 periods of 5 minutes each.\\
(b) Find the probability that in exactly 3 of these periods there were no calls.
On another occasion Indre took 1 break of 5 minutes and 1 break of 15 minutes.\\
(c) Find the probability that Indre missed exactly 1 call in each of these 2 breaks.
\hfill \mbox{\textit{Edexcel FS1 2019 Q2 [8]}}