| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Conditional probability with Poisson |
| Difficulty | Challenging +1.2 Part (a) is routine Poisson calculation with scaled mean (0.15×50=7.5). Part (b) requires multi-step reasoning: recognizing that 'at least 2 samples with no particles' follows binomial distribution, where p=P(X=0)=e^(-0.15m), then using complement probability and tables to work backwards to find m. This inverse problem with nested distributions elevates it above standard S2 questions but remains within typical exam scope. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(X \sim Po(7.5)\) | B1 | |
| (i) \(P(X = 10) = [0.8622 - 0.7764 = \frac{e^{-7.5}(7.5)^{10}}{10!}] = 0.0858...\) awrt 0.0858 | B1 | 2nd B1 awrt 0.0858 [calc = 0.0858303...] |
| (ii) \(P(6_{,n} X_{,n} 11) = P(X_{,n} 11) - P(X_{,n} 5) = [0.9208 - 0.2414]\) | M1 | |
| \(= 0.6794\) awrt 0.679 | A1 | |
| (b) \(Y =\) number of samples that contain 0 particles | M1 | |
| \(Y \sim B(12, p)\) or \(B(12, e^{-0.15m})\) or \(B(12, e^{-4})\) | M1 | |
| \([P(Y ... 2)] = 1 - P(Y_{,n} 1) = 0.1184\) | M1 | |
| \(P(Y_{,n} 1) = 0.8816 →\) from tables \([p =] 0.05\) | A1 | 1st A1 0.05(seen) |
| \(S =\) number of particles per \(m\) millilitres | M1 | |
| \(S \sim Po(0.15m)\) | M1 | |
| \(P(S = 0) = 0.05\) or \(e^{-0.15m} = "0.05"\) | M1 | |
| \(-0.15m = \ln(0.05) → m = 19.9715...\) awrt 20.0 | A1 | 2nd A1 Allow 20 or awrt 20.0 Allow trial and error to solve their equation |
| (c) | 1st B1 writing or using Po(7.5) May be implied by a correct probability. 1st A1 0.0858 [calc = 0.0858303...]. M1 writing or using P(X_{,n} 11) - P(X_{,n} 5). A1 awrt 0.0679 [calc = 0.06793222...]. 1st M1 writing or using B(12, p) Allow Binomial with n = 12 or B(12,...) May be implied by 0.05. 2nd M1 for 1 - P(Y_{,n} 1) = 0.1184 (or better) or P(Y_{,n} 1) = 0.8816 oe eg (1 - p)^12 + 12p(1 - p)^11 = 0.8816 Implied by 0.05. 3rd M1 writing or using Po(0.15m) May be implied by e^{-0.15m}. 4th M1 If their p (0 < p < 1) for an equation of the form e^{-0.15m} = "0.05" (allow e^{-1} = "0.05"). Allow 0.15m = 3. 2nd A1 Allow 20 or awrt 20.0 Allow trial and error to solve their equation |
**(a)** $X \sim Po(7.5)$ | B1 |
**(i)** $P(X = 10) = [0.8622 - 0.7764 = \frac{e^{-7.5}(7.5)^{10}}{10!}] = 0.0858...$ awrt **0.0858** | B1 | 2nd B1 awrt 0.0858 [calc = 0.0858303...]
**(ii)** $P(6_{,n} X_{,n} 11) = P(X_{,n} 11) - P(X_{,n} 5) = [0.9208 - 0.2414]$ | M1 |
$= 0.6794$ awrt **0.679** | A1 |
**(b)** $Y =$ number of samples that contain 0 particles | M1 |
$Y \sim B(12, p)$ or $B(12, e^{-0.15m})$ or $B(12, e^{-4})$ | M1 |
$[P(Y ... 2)] = 1 - P(Y_{,n} 1) = 0.1184$ | M1 |
$P(Y_{,n} 1) = 0.8816 →$ from tables $[p =] 0.05$ | A1 | 1st A1 0.05(seen)
$S =$ number of particles per $m$ millilitres | M1 |
$S \sim Po(0.15m)$ | M1 |
$P(S = 0) = 0.05$ or $e^{-0.15m} = "0.05"$ | M1 |
$-0.15m = \ln(0.05) → m = 19.9715...$ awrt **20.0** | A1 | 2nd A1 Allow 20 or awrt 20.0 Allow trial and error to solve their equation
**(c)** | | 1st B1 writing or using Po(7.5) May be implied by a correct probability. 1st A1 0.0858 [calc = 0.0858303...]. M1 writing or using P(X_{,n} 11) - P(X_{,n} 5). A1 awrt 0.0679 [calc = 0.06793222...]. 1st M1 writing or using B(12, p) Allow Binomial with n = 12 or B(12,...) May be implied by 0.05. 2nd M1 for 1 - P(Y_{,n} 1) = 0.1184 (or better) or P(Y_{,n} 1) = 0.8816 oe eg (1 - p)^12 + 12p(1 - p)^11 = 0.8816 Implied by 0.05. 3rd M1 writing or using Po(0.15m) May be implied by e^{-0.15m}. 4th M1 If their p (0 < p < 1) for an equation of the form e^{-0.15m} = "0.05" (allow e^{-1} = "0.05"). Allow 0.15m = 3. 2nd A1 Allow 20 or awrt 20.0 Allow trial and error to solve their equation
**[10 marks]**
---\begin{enumerate}
\item The number of particles per millilitre in a solution is modelled by a Poisson distribution with mean 0.15
\end{enumerate}
A randomly selected 50 millilitre sample of the solution is taken.\\
(a) Find the probability that\\
(i) exactly 10 particles are found,\\
(ii) between 6 and 11 particles (inclusive) are found.
Petra takes 12 independent samples of $m$ millilitres of the solution.\\
The probability that at least 2 of these samples contain no particles is 0.1184\\
(b) Using the Statistical Tables provided, find the value of $m$
\hfill \mbox{\textit{Edexcel S2 2022 Q5 [10]}}