| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | October |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Moderate -0.8 This is a straightforward S2 question testing standard Poisson distribution knowledge: stating conditions (textbook recall), basic probability calculations with given parameters, a routine hypothesis test, and standard normal approximation. All parts follow predictable patterns with no novel problem-solving required, making it easier than average for A-level. |
| Spec | 2.04d Normal approximation to binomial2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail5.02i 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 |
| Common Spotted-orchids occur singly/randomly/independently | B1 | One of the given reasons; no context needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S \sim Po(4.5)\) | — | |
| \(P(S=6) = \frac{e^{-4.5}4.5^6}{6!}\) or \(P(S \leq 6) - P(S \leq 5)\) | M1 | For \(\frac{e^{-\lambda}\lambda^6}{6!}\) with any \(\lambda\), or writing/using \(P(S \leq 6) - P(S \leq 5)\) |
| \(= 0.1281...\) | A1 | awrt 0.128 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(4 < S < 10) = P(S \leq 9) - P(S \leq 4)\) or \(0.9829 - 0.5321\) | M1 | Writing or using \(P(S \leq 9) - P(S \leq 4)\) |
| \(= 0.4508\) | A1 | awrt 0.451 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \lambda = 9\), \(H_1: \lambda > 9\) | B1 | Both hypotheses correct; must be in terms of \(\lambda\) or \(\mu\); allow 4.5 instead of 9 |
| \(M \sim Po(9)\), \(P(M \geq 11) = 1 - P(M \leq 10)\) or \(P(M \geq 15) = 0.0415\) | M1 | Writing or using \(Po(9)\) and \(1-P(M \leq 10)\) or \(P(M \geq 15) = 0.0415\); implied by correct CR |
| \(= 0.294\) or CR \(M \geq 15\) | A1 | 0.3 or 0.29... or better (0.2940...) or \(M \geq 15\) oe |
| Accept \(H_0\) or insignificant; 11 does not lie in critical region | dM1 | Dep on M1 A1; correct statement, no context needed |
| There is insufficient evidence to support Juan's belief | A1 | Allow claim instead of belief; \(\lambda < 9\) gets A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T \sim N(90, 90)\) | B1 | Writing or using \(N(90, 90)\) |
| \(P(T < 70) = P\left(Z < \pm\left(\frac{69.5 - 90}{\sqrt{90}}\right)\right)\) or \(P(Z < \pm 2.160...)\) | M1 | Standardising with 68.5, 69.5, or 70.5 and mean and sd; awrt 2.16 |
| \(= 0.0154\) | A1 | awrt 0.0154; NB Poisson gives 0.01275... |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V \sim Po(200 \times 0.012) = Po(2.4)\) | M1 | Writing or using \(Po(200 \times 0.012)\); allow \(Po(200 \times \text{"their d"})\) |
| \(P(V=0) + P(V=1) = e^{-2.4}(1 + 2.4)\) | dM1 | Dependent on using Poisson; for using/writing \(P(V=0)+P(V=1)\) or \(e^{-\lambda}(1+\lambda)\) or \(P(V \leq 1)\) oe |
| \(= 0.30844...\) | A1 | awrt 0.308; NB Binomial gives 0.3066 |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Common Spotted-orchids occur singly/randomly/independently | B1 | One of the given reasons; no context needed |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S \sim Po(4.5)$ | — | |
| $P(S=6) = \frac{e^{-4.5}4.5^6}{6!}$ or $P(S \leq 6) - P(S \leq 5)$ | M1 | For $\frac{e^{-\lambda}\lambda^6}{6!}$ with any $\lambda$, or writing/using $P(S \leq 6) - P(S \leq 5)$ |
| $= 0.1281...$ | A1 | awrt 0.128 |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(4 < S < 10) = P(S \leq 9) - P(S \leq 4)$ or $0.9829 - 0.5321$ | M1 | Writing or using $P(S \leq 9) - P(S \leq 4)$ |
| $= 0.4508$ | A1 | awrt 0.451 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 9$, $H_1: \lambda > 9$ | B1 | Both hypotheses correct; must be in terms of $\lambda$ or $\mu$; allow 4.5 instead of 9 |
| $M \sim Po(9)$, $P(M \geq 11) = 1 - P(M \leq 10)$ or $P(M \geq 15) = 0.0415$ | M1 | Writing or using $Po(9)$ and $1-P(M \leq 10)$ or $P(M \geq 15) = 0.0415$; implied by correct CR |
| $= 0.294$ or CR $M \geq 15$ | A1 | 0.3 or 0.29... or better (0.2940...) or $M \geq 15$ oe |
| Accept $H_0$ or insignificant; 11 does not lie in critical region | dM1 | Dep on M1 A1; correct statement, no context needed |
| There is insufficient evidence to support **Juan's belief** | A1 | Allow claim instead of belief; $\lambda < 9$ gets A0 |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T \sim N(90, 90)$ | B1 | Writing or using $N(90, 90)$ |
| $P(T < 70) = P\left(Z < \pm\left(\frac{69.5 - 90}{\sqrt{90}}\right)\right)$ or $P(Z < \pm 2.160...)$ | M1 | Standardising with 68.5, 69.5, or 70.5 and mean and sd; awrt 2.16 |
| $= 0.0154$ | A1 | awrt 0.0154; NB Poisson gives 0.01275... |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V \sim Po(200 \times 0.012) = Po(2.4)$ | M1 | Writing or using $Po(200 \times 0.012)$; allow $Po(200 \times \text{"their d"})$ |
| $P(V=0) + P(V=1) = e^{-2.4}(1 + 2.4)$ | dM1 | Dependent on using Poisson; for using/writing $P(V=0)+P(V=1)$ or $e^{-\lambda}(1+\lambda)$ or $P(V \leq 1)$ oe |
| $= 0.30844...$ | A1 | awrt 0.308; NB Binomial gives 0.3066 |
---4. In a peat bog, Common Spotted-orchids occur at a mean rate of 4.5 per $\mathrm { m } ^ { 2 }$
\begin{enumerate}[label=(\alph*)]
\item Give an assumption, not already stated, that is required for the number of Common Spotted-orchids per $\mathrm { m } ^ { 2 }$ of the peat bog to follow a Poisson distribution.\\
(1)
Given that the number of Common Spotted-orchids in $1 \mathrm {~m} ^ { 2 }$ of the peat bog can be modelled by a Poisson distribution,
\item find the probability that in a randomly selected $1 \mathrm {~m} ^ { 2 }$ of the peat bog
\begin{enumerate}[label=(\roman*)]
\item there are exactly 6 Common Spotted-orchids,
\item there are fewer than 10 but more than 4 Common Spotted-orchids.\\
(4)
Juan believes that by introducing a new management scheme the number of Common Spotted-orchids in the peat bog will increase. After three years under the new management scheme, a randomly selected $2 \mathrm {~m} ^ { 2 }$ of the peat bog contains 11 Common Spotted-orchids.
\end{enumerate}\item Using a $5 \%$ significance level assess Juan's belief. State your hypotheses clearly.
Assuming that in the peat bog, Common Spotted-orchids still occur at a mean rate of 4.5 per $\mathrm { m } ^ { 2 }$
\item use a normal approximation to find the probability that in a randomly selected $20 \mathrm {~m} ^ { 2 }$ of the peat bog there are fewer than 70 Common Spotted-orchids.
Following a period of dry weather, the probability that there are fewer than 70 Common Spotted-orchids in a randomly selected $20 \mathrm {~m} ^ { 2 }$ of the peat bog is 0.012
A random sample of 200 non-overlapping $20 \mathrm {~m} ^ { 2 }$ areas of the peat bog is taken.
\item Using a suitable approximation, calculate the probability that at most 1 of these areas contains fewer than 70 Common Spotted-orchids.
\includegraphics[max width=\textwidth, alt={}, center]{3a781851-e2cc-4379-8b8c-abb3060a6019-15_2255_50_314_34}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2020 Q4 [16]}}