| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| 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 Poisson distribution calculations, a Poisson approximation to binomial, a hypothesis test for Poisson mean, and working backwards from a probability inequality. While each individual technique is standard for FS1, the combination of five parts testing different aspects of Poisson modeling, plus the need to combine two independent Poisson processes in part (d), makes this more demanding than a typical A-level question but still within standard Further Maths scope. |
| 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 |
|---|---|---|
| \(W \sim \text{Po}(11.2)\) and \(P(W \ldots 19) = 1 - P(W \text{ ,, } 18)\) or suitable 3sf probs | M1 | For using model Po(11.2) implied by sight of: 0.02077... or 0.9889.. or 0.9792.. |
| \(P(W \ldots 19) = 0.020776\ldots\) awrt \(\mathbf{0.021}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \([S = \#\) calls per day, \(S \sim \text{Po}(0.4)]\) \(P(S>1) = 0.061551\ldots\) awrt 0.0616 | B1 | |
| \(X \sim \text{B}(250, \text{"0.061551..."})\) | M1 | Setting up new model B(250, "0.0616") [condone B("0.0616", 250)] |
| \(Y \sim \text{Po}(\text{"15.3879..."})\) [Accept Po(15.4) or better] or suitable 3sf probs | M1 | Seeing model Po(their \(np\)) implied by sight of: 0.1475.. or 0.89975 or 0.8524... |
| \(= 0.14751\ldots\) awrt \(\mathbf{0.148}\) | A1 | if no approximation used (and 1st M1 not seen) answer of awrt 0.140 could get B1M1M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \lambda = 16.8 \quad H_1: \lambda < 16.8\) | B1 | Both hypotheses correct using \(\lambda\) or \(\mu\) and 16.8 or 0.4 [Accept ans to \(0.4\times42\)] |
| \(U \sim \text{Po}(16.8)\) | B1 | Realising Po(16.8) needs to be used |
| \(P(U \text{ ,, } 8) = 0.014\) | M1 | For 0.014 or better (0.0141..) or CR \(X\) ,, 9; must be CR and not probability. [Allow CR \(X\) ,, 10 with probability \(P(X\text{ ,, }10)=0.054\) or better] |
| \([0.014 < 0.05\) or there is sufficient evidence to reject \(H_0]\) There is sufficient evidence at the 5% level of significance that the number of calls received per day is lower in winter | A1 | Indep of 1st B1 (must see 2nd B1 and M1 scored) for correct inference in context |
| Answer | Marks | Guidance |
|---|---|---|
| \(C \sim \text{Po}(0.4\times n + 0.2\times n)[= \text{Po}(0.6n)]\) or \(D \sim \text{B}(n, e^{-0.6}\) or awrt 0.549) | M1 | Selecting suitable model |
| \(e^{-0.6n} < 0.001\) or \(-0.6n < \ln(0.001)\) or \(n > 11.5\ldots\) | M1 | For correct inequality or equality involving \(n\) [Condone slips in solving]. Allow MR i.e. misread of 0.01 for 0.001 to score M1M1A0 |
| \(n = \mathbf{12}\) | A1 | \(n=12\) cao. Correct answer with no incorrect working scores 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| The rate of calls per day is constant or the number of calls occurring in non-overlapping time intervals is independent or number of calls per day is independent | B1 | Allow equivalent statements. Underlined words required |
## Question 3:
### Part (a):
| $W \sim \text{Po}(11.2)$ and $P(W \ldots 19) = 1 - P(W \text{ ,, } 18)$ or suitable 3sf probs | M1 | For using model Po(11.2) implied by sight of: 0.02077... or 0.9889.. or 0.9792.. |
| $P(W \ldots 19) = 0.020776\ldots$ awrt $\mathbf{0.021}$ | A1 | |
### Part (b):
| $[S = \#$ calls per day, $S \sim \text{Po}(0.4)]$ $P(S>1) = 0.061551\ldots$ awrt 0.0616 | B1 | |
| $X \sim \text{B}(250, \text{"0.061551..."})$ | M1 | Setting up new model B(250, "0.0616") [condone B("0.0616", 250)] |
| $Y \sim \text{Po}(\text{"15.3879..."})$ [Accept Po(15.4) or better] or suitable 3sf probs | M1 | Seeing model Po(their $np$) implied by sight of: 0.1475.. or 0.89975 or 0.8524... |
| $= 0.14751\ldots$ awrt $\mathbf{0.148}$ | A1 | if no approximation used (and 1st M1 not seen) answer of awrt 0.140 could get B1M1M0A0 |
### Part (c):
| $H_0: \lambda = 16.8 \quad H_1: \lambda < 16.8$ | B1 | Both hypotheses correct using $\lambda$ or $\mu$ and 16.8 or 0.4 [Accept ans to $0.4\times42$] |
| $U \sim \text{Po}(16.8)$ | B1 | Realising Po(16.8) needs to be used |
| $P(U \text{ ,, } 8) = 0.014$ | M1 | For 0.014 or better (0.0141..) or CR $X$ ,, 9; must be CR and not probability. [Allow CR $X$ ,, 10 with probability $P(X\text{ ,, }10)=0.054$ or better] |
| $[0.014 < 0.05$ or there is sufficient evidence to reject $H_0]$ There is sufficient evidence at the 5% level of significance that the number of calls received per day is lower in winter | A1 | Indep of 1st B1 (must see 2nd B1 and M1 scored) for correct inference in context |
### Part (d):
| $C \sim \text{Po}(0.4\times n + 0.2\times n)[= \text{Po}(0.6n)]$ or $D \sim \text{B}(n, e^{-0.6}$ or awrt 0.549) | M1 | Selecting suitable model |
| $e^{-0.6n} < 0.001$ or $-0.6n < \ln(0.001)$ or $n > 11.5\ldots$ | M1 | For correct inequality or equality involving $n$ [Condone slips in solving]. Allow MR i.e. misread of 0.01 for 0.001 to score M1M1A0 |
| $n = \mathbf{12}$ | A1 | $n=12$ cao. Correct answer with no incorrect working scores 3/3 |
### Part (e):
| The rate of calls per day is constant **or** the number of calls occurring in non-overlapping time intervals is independent **or** number of calls per day is independent | B1 | Allow equivalent statements. Underlined words required |
---\begin{enumerate}
\item During the summer, mountain rescue team $A$ receives calls for help randomly with a rate of 0.4 per day.\\
(a) Find the probability that during the summer, mountain rescue team $A$ receives at least 19 calls for help in 28 randomly selected days.
\end{enumerate}
The leader of mountain rescue team $A$ randomly selects 250 summer days from the last few years.\\
She records the number of calls for help received on each of these days.\\
(b) Using a Poisson approximation, estimate the probability of the leader finding at least 20 of these days when more than 1 call for help was received by mountain rescue team $A$.
Mountain rescue team $A$ believes that the number of calls for help per day is lower in the winter than in the summer. The number of calls for help received in 42 randomly selected winter days is 8\\
(c) Use a suitable test, at the $5 \%$ level of significance, to assess whether or not there is evidence that the number of calls for help per day is lower in the winter than in the summer. State your hypotheses clearly.
During the summer, mountain rescue team $B$ receives calls for help randomly with a rate of 0.2 per day, independently of calls to mountain rescue team $A$.
The random variable $C$ is the total number of calls for help received by mountain rescue teams $A$ and $B$ during a period of $n$ days in the summer.\\
On a Monday in the summer, mountain rescue teams $A$ and $B$ each receive a call for help.
Given that over the next $n$ days $\mathrm { P } ( C = 0 ) < 0.001$\\
(d) calculate the minimum value of $n$\\
(e) Write down an assumption that needs to be made for the model to be appropriate.
\hfill \mbox{\textit{Edexcel FS1 2022 Q3 [14]}}