| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with binomial combination |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths Statistics question requiring Poisson distribution calculations, binomial-Poisson combination, Poisson approximation to binomial, and hypothesis testing. While it involves several techniques and careful parameter scaling (10 min → 30 min → 4 hours), each part follows standard procedures without requiring novel insight. The combination of topics and extended working places it moderately above average difficulty for A-level, but it's a straightforward application of learned methods for Further Maths students. |
| 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 |
| \(C \sim \text{Poisson}(3.75)\) | M1 | For calculating mean and setting up correct model; Poisson may be implied by 0.8883 or \(1 - \text{awrt } 0.1117\); must see 3.75 or \(1.25 \times 3\) |
| \(P(C \geqslant 2) = 0.88829\ldots\) awrt 0.8883 | A1*cso | \(P(C \geqslant 2) =\) awrt 0.8883 or \(1 -\) awrt \(0.1117 = 0.888\); must see \(P(C \geqslant 2)\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(D \sim B(6,\ \text{"0.888"})\) | M1 | Setting up new model using answer to (a); implied by correct answer |
| \(P(D \leqslant 3) = 0.02163\ldots\) awrt 0.0216/0.0215 | A1 | awrt 0.0216 or awrt 0.0215 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(C = 8) = 0.02281\ldots\) | B1 | awrt 0.0228 |
| \(E \sim B(150,\ \text{"0.02281"}) \Rightarrow \text{mean} = 150 \times \text{"0.02281"} = [3.4215\ldots]\) | M1 | Setting up new model \(B(150, \text{"0.0228"})\) and using \(np\) (working seen if incorrect) |
| \(E \sim \text{Po}(\text{"3.4215"}) \Rightarrow P(E \geqslant 3) = [1 - P(E \leqslant 2)]\) | M1 | Using model \(\text{Po}(np)\); must be clearly stated and \(P(E \geqslant 3)\) oe seen |
| \(= 0.664\) | A1*cso | Only award if previous 3 marks awarded and 0.664 stated; NB \(B(150, 0.02281)\) gives 0.668 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The number of periods is large and the probability of receiving 8 calls in 30 minutes is small | B1 | Idea that \(n = 150\) is large and \(p = 0.022\ldots\) is small |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda = 30 \qquad H_1: \lambda \neq 30\) | B1 | Both hypotheses correct using \(\lambda\) or \(\mu\); allow 1.25 or 3.75 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim \text{Po}(30)\) | B1 | Realising \(\text{Po}(30)\) needed; implied by correct answer or \(P(X = 40) = 0.0139\ldots\) |
| \(P(X \geqslant 40) = 1 - P(X \leqslant 39)\) | M1 | Writing/using \(1 - P(X \leqslant 39)\); or if CR method, \(P(X \geqslant 42) = 0.0221\ldots\) |
| \(= 0.04625\ldots\) | A1 | \(0.04\ldots\) or awrt 0.05; or CR \(X \geqslant 42\) (must be CR, not probability) |
| \(0.046\ldots > 0.025\), no evidence to reject \(H_0\); insufficient evidence at 5% level that the number of calls received is different on a Saturday | A1 | Fully correct solution and correct inference in context; "calls" required; if probability given, then award Cr \(X \geqslant 40\) M1A1A0 |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $C \sim \text{Poisson}(3.75)$ | M1 | For calculating mean and setting up correct model; Poisson may be implied by 0.8883 or $1 - \text{awrt } 0.1117$; must see 3.75 or $1.25 \times 3$ |
| $P(C \geqslant 2) = 0.88829\ldots$ awrt 0.8883 | A1*cso | $P(C \geqslant 2) =$ awrt 0.8883 or $1 -$ awrt $0.1117 = 0.888$; must see $P(C \geqslant 2)$ oe |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $D \sim B(6,\ \text{"0.888"})$ | M1 | Setting up new model using answer to (a); implied by correct answer |
| $P(D \leqslant 3) = 0.02163\ldots$ awrt 0.0216/0.0215 | A1 | awrt 0.0216 or awrt 0.0215 |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(C = 8) = 0.02281\ldots$ | B1 | awrt 0.0228 |
| $E \sim B(150,\ \text{"0.02281"}) \Rightarrow \text{mean} = 150 \times \text{"0.02281"} = [3.4215\ldots]$ | M1 | Setting up new model $B(150, \text{"0.0228"})$ and using $np$ (working seen if incorrect) |
| $E \sim \text{Po}(\text{"3.4215"}) \Rightarrow P(E \geqslant 3) = [1 - P(E \leqslant 2)]$ | M1 | Using model $\text{Po}(np)$; must be clearly stated and $P(E \geqslant 3)$ oe seen |
| $= 0.664$ | A1*cso | Only award if previous 3 marks awarded and 0.664 stated; NB $B(150, 0.02281)$ gives 0.668 |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| The number of periods is large and the probability of receiving 8 calls in 30 minutes is small | B1 | Idea that $n = 150$ is large and $p = 0.022\ldots$ is small |
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 30 \qquad H_1: \lambda \neq 30$ | B1 | Both hypotheses correct using $\lambda$ or $\mu$; allow 1.25 or 3.75 |
## Part (f)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim \text{Po}(30)$ | B1 | Realising $\text{Po}(30)$ needed; implied by correct answer or $P(X = 40) = 0.0139\ldots$ |
| $P(X \geqslant 40) = 1 - P(X \leqslant 39)$ | M1 | Writing/using $1 - P(X \leqslant 39)$; or if CR method, $P(X \geqslant 42) = 0.0221\ldots$ |
| $= 0.04625\ldots$ | A1 | $0.04\ldots$ or awrt 0.05; or CR $X \geqslant 42$ (must be CR, not probability) |
| $0.046\ldots > 0.025$, no evidence to reject $H_0$; insufficient evidence at 5% level that the number of **calls** received is different on a Saturday | A1 | Fully correct solution and correct inference in context; "calls" required; if probability given, then award Cr $X \geqslant 40$ M1A1A0 |\begin{enumerate}
\item On a weekday, a garage receives telephone calls randomly, at a mean rate of 1.25 per 10 minutes.\\
(a) Show that the probability that on a weekday at least 2 calls are received by the garage in a 30 -minute period is 0.888 to 3 decimal places.\\
(b) Calculate the probability that at least 2 calls are received by the garage in fewer than 4 out of 6 randomly selected, non-overlapping 30-minute periods on a weekday.
\end{enumerate}
The manager of the garage randomly selects 150 non-overlapping 30-minute periods on weekdays.\\
She records the number of calls received in each of these 30-minute periods.\\
(c) Using a Poisson approximation show that the probability of the manager finding at least 3 of these 30 -minute periods when exactly 8 calls are received by the garage is 0.664 to 3 significant figures.\\
(d) Explain why the Poisson approximation may be reasonable in this case.
The manager of the garage decides to test whether the number of calls received on a Saturday is different from the number of calls received on a weekday. She selects a Saturday at random and records the number of telephone calls received by the garage in the first 4 hours.\\
(e) Write down the hypotheses for this test.
The manager found that there had been 40 telephone calls received by the garage in the first 4 hours.\\
(f) Carry out the test using a $5 \%$ level of significance.
\hfill \mbox{\textit{Edexcel FS1 2021 Q2 [14]}}