| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.3 Part (a) is straightforward Poisson probability calculation with λ=3. Part (b) requires solving P(X=0)=e^(-0.6t)<0.16 using logarithms, which is a standard technique. Part (c) is a hypothesis test comparing observed data to a Poisson distribution with combined rate 5.6, requiring calculation of p-value and interpretation. While multi-part and requiring several techniques, all components are standard FS1 procedures with no novel insight needed—slightly easier than average A-level difficulty. |
| 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 modelling5.02m Poisson: mean = variance = lambda5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim \text{Po}(3)\) | M1 | Writing or using Po(3) |
| \(P(X = 4) = 0.1680...\) | A1 | awrt 0.168 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^{-0.6 \times t} < 0.16\) oe | M1 | Forming correct equation from information given. Condone \(e^{-0.6t} = 0.16\) or finding \(P(X=0)\) for \([t=3.1]\ 0.155...\) and \([t=3]\ 0.165...\) or \(P(X=0)\) for \([\lambda=1.84]\ 0.158...\) and \([\lambda=1.83]\ 0.1604...\) |
| \(-0.6 \times t < \ln 0.16\) | dM1 | Dependent on 1st method mark. Correct method to solve inequality/equation. Or \([t=3.05]\ 0.1604\) or \([\lambda=1.835]\ 0.159...\) |
| \([t > 3.054...]\) \(\quad t = 3.1\) | A1 | An answer of 3.1 gains 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda = 1.4 \quad H_1: \lambda > 1.4\) | B1 | Both hypotheses in terms of \(\lambda\) or \(\mu\). Allow 5.6 instead of 1.4 |
| \(J \sim \text{Po}(5.6)\) | B1 | Writing or using Po(5.6) |
| \(P(J \geqslant 12) = 1 - P(J \leqslant 11)\) | M1 | For writing or using \(1 - P(J \leqslant 11)\). Implied by a correct probability or CR. Allow \(P(J \leqslant 10) = \text{awrt}\ 0.972\) and \(P(J \leqslant 9) = \text{awrt}\ 0.941\) |
| \(= 1 - 0.9875...\) | ||
| \(= 0.01(248...)\) | A1 | 0.01 or better (allow truncation e.g. 0.0124). NB Allow M1A1 if \(P(J \leqslant 11) = 0.9875...\) is written on its own |
| Method 2: \(P(J \geqslant 11) = \text{awrt}\ 0.0282\) and \(P(J \geqslant 10) = \text{awrt}\ 0.0591...\); \(J \geqslant 11\) | A1 | |
| \(0.01(24) < 0.05\) or \(12 > 11\) or 12 is in the critical region or 12 is significant or Reject \(H_0\). There is evidence at the 5% level of significance that the rate of fish caught may have increased. | A1 | Independent of hypotheses. Correct conclusion based on their probability with 0.05, conclusion in context (bold words). Do not accept contradicting statements. |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim \text{Po}(3)$ | M1 | Writing or using Po(3) |
| $P(X = 4) = 0.1680...$ | A1 | awrt 0.168 |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{-0.6 \times t} < 0.16$ oe | M1 | Forming correct equation from information given. Condone $e^{-0.6t} = 0.16$ or finding $P(X=0)$ for $[t=3.1]\ 0.155...$ and $[t=3]\ 0.165...$ or $P(X=0)$ for $[\lambda=1.84]\ 0.158...$ and $[\lambda=1.83]\ 0.1604...$ |
| $-0.6 \times t < \ln 0.16$ | dM1 | Dependent on 1st method mark. Correct method to solve inequality/equation. Or $[t=3.05]\ 0.1604$ or $[\lambda=1.835]\ 0.159...$ |
| $[t > 3.054...]$ $\quad t = 3.1$ | A1 | An answer of 3.1 gains 3/3 |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 1.4 \quad H_1: \lambda > 1.4$ | B1 | Both hypotheses in terms of $\lambda$ or $\mu$. Allow 5.6 instead of 1.4 |
| $J \sim \text{Po}(5.6)$ | B1 | Writing or using Po(5.6) |
| $P(J \geqslant 12) = 1 - P(J \leqslant 11)$ | M1 | For writing or using $1 - P(J \leqslant 11)$. Implied by a correct probability or CR. Allow $P(J \leqslant 10) = \text{awrt}\ 0.972$ and $P(J \leqslant 9) = \text{awrt}\ 0.941$ |
| $= 1 - 0.9875...$ | | |
| $= 0.01(248...)$ | A1 | 0.01 or better (allow truncation e.g. 0.0124). **NB** Allow M1A1 if $P(J \leqslant 11) = 0.9875...$ is written on its own |
| Method 2: $P(J \geqslant 11) = \text{awrt}\ 0.0282$ and $P(J \geqslant 10) = \text{awrt}\ 0.0591...$; $J \geqslant 11$ | A1 | |
| $0.01(24) < 0.05$ or $12 > 11$ or 12 is in the critical region or 12 is significant or Reject $H_0$. There is evidence at the 5% level of significance that the **rate** of fish caught may have **increased**. | A1 | Independent of hypotheses. Correct conclusion based on their probability with 0.05, conclusion in context (bold words). Do not accept contradicting statements. |
---\begin{enumerate}
\item Xena catches fish at random, at a constant rate of 0.6 per hour.\\
(a) Find the probability that Xena catches exactly 4 fish in a 5 -hour period.
\end{enumerate}
The probability of Xena catching no fish in a period of $t$ hours is less than 0.16\\
(b) Find the minimum value of $t$, giving your answer to one decimal place.
Independently of Xena, Zion catches fish at random with a mean rate of 0.8 per hour.\\
Xena and Zion try using new bait to catch fish. The number of fish caught in total by Xena and Zion after using the new bait, in a randomly selected 4-hour period, is 12\\
(c) Use a suitable test to determine, at the $5 \%$ level of significance, whether or not there is evidence that the rate at which fish are caught has increased after using the new bait. State your hypotheses clearly and the $p$-value used in your test.
\hfill \mbox{\textit{Edexcel FS1 AS 2022 Q2 [10]}}