| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with binomial combination |
| Difficulty | Standard +0.8 This S2 question requires three distinct techniques: basic Poisson probability (routine), binomial probability with Poisson parameter scaling (moderate), and reverse normal approximation requiring standardization and solving for the parameter (challenging for A-level). Part (c) especially demands careful manipulation of the continuity correction and z-score formula, which is non-routine. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F\) represents number of flaws per 50 m \(\Rightarrow F \sim \text{Po}(2)\) | M1 | |
| \(P(F = 5) = 0.9834 - 0.9473\) or \(\frac{e^{-2}2^5}{5!}\) | M1 | (2) |
| \(= 0.0361\) | A1 | |
| \(G\) represents number of flaws per 200 m \(\Rightarrow G \sim \text{Po}(8)\) | B1 | |
| \(P(G < 7) = P(G \leq 6) = 0.3134\) | B1 | |
| \([R = \text{number of 200 m rolls containing fewer than 7 flaws}] \quad R \sim B(4, 0.3134)\) | M1A1ft | |
| \(P(R = 1) = C_1^4 \times 0.3134 \times (1-0.3134)^3 = 0.40576...\) | M1 A1 | awrt 0.406 |
| \(N\) represents number of flaws in a \(x\) m roll \(\Rightarrow N \sim \text{Po}(2)\) | M1, M1 A1 | |
| \(P(N < 26) = P\left(\frac{25.5 - \lambda}{\sqrt{\lambda}}\right)\) | M1, M1 A1 | \(\pm 0.5\), standardise |
| \(\frac{25.5 - \lambda}{\sqrt{\lambda}} = 0.1\) gives \(\lambda + 0.1\sqrt{\lambda} - 25.5 = 0\) | B1 | |
| \(\sqrt{\lambda} = \frac{-0.1 \pm \sqrt{0.1^2 + 4 \times 25.5}}{2}\) | dM1 | |
| \(\left[\sqrt{\lambda} = 5\right]\) so \(\lambda = 25\) | A1 | |
| \(x = \frac{25}{2} \times 50\) so \(x = 625\) m | dM1, A1 | (8) |
| [16] |
| Answer | Marks | Guidance |
|---|---|---|
| Item | Guidance | |
| (a) | M1: Writing \(P(X \leq 5) - P(X \leq 4)\) or \(\frac{e^{-\lambda}\lambda^5}{5!}\) (any value of \(\lambda\)) | A1: awrt 0.0361 |
| (b) | 1st B1: Writing or using Po(8) | 2nd B1: awrt 0.313 (calc gives 0.3133742...) |
| 1st M1: Recognize Binomial | 1st A1ft: writing \(B(4, \text{'their } 0.313')\) May be \(\Rightarrow\) by next line | |
| 2nd dM1 (dep. on 1st M1): \(C_1^4 \times \text{'their } 0.3134 \times (1-\text{'their } 0.3134)^3\) | 2nd A1: awrt 0.406 | |
| (c) | 1st M1: continuity correction used. Either 25.5 or 26.5 | 2nd M1: standardising using their \(\lambda\) and \(\sqrt{\lambda}\) for mean and sd. Any letter may be used or \(\frac{\lambda}{25}\) etc |
| 1st A1: \(\frac{25.5 - \lambda}{\sqrt{\lambda}} = z\) where \(0 < z < 0.5\). May be implied by their correct quadratic (25.5 req'd) | ||
| B1: 0.1 (calc 0.09992...) used as their z value in an equation. Allow e.g. \(\frac{26-\mu}{\sigma} = 0.1\) | ||
| 3rd dM1 (dep on 2nd M1): some attempt at solving their 3TQ \(= \frac{-b \pm \sqrt{b^2+4ac}}{2a}\) | 2nd A1: 25 (o.e.) | |
| 4th dM1 (dep on 3rd M1): \(\text{their } \frac{25}{2} \times 50\) (If using \(\frac{\lambda}{25}\) award when \(x = \) ...) | 3rd A1: awrt 625 |
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F$ represents number of flaws per 50 m $\Rightarrow F \sim \text{Po}(2)$ | M1 | |
| $P(F = 5) = 0.9834 - 0.9473$ or $\frac{e^{-2}2^5}{5!}$ | M1 | (2) |
| $= 0.0361$ | A1 | |
| $G$ represents number of flaws per 200 m $\Rightarrow G \sim \text{Po}(8)$ | B1 | |
| $P(G < 7) = P(G \leq 6) = 0.3134$ | B1 | |
| $[R = \text{number of 200 m rolls containing fewer than 7 flaws}] \quad R \sim B(4, 0.3134)$ | M1A1ft | |
| $P(R = 1) = C_1^4 \times 0.3134 \times (1-0.3134)^3 = 0.40576...$ | M1 A1 | awrt 0.406 | (6) |
| $N$ represents number of flaws in a $x$ m roll $\Rightarrow N \sim \text{Po}(2)$ | M1, M1 A1 | |
| $P(N < 26) = P\left(\frac{25.5 - \lambda}{\sqrt{\lambda}}\right)$ | M1, M1 A1 | $\pm 0.5$, standardise | |
| $\frac{25.5 - \lambda}{\sqrt{\lambda}} = 0.1$ gives $\lambda + 0.1\sqrt{\lambda} - 25.5 = 0$ | B1 | |
| $\sqrt{\lambda} = \frac{-0.1 \pm \sqrt{0.1^2 + 4 \times 25.5}}{2}$ | dM1 | |
| $\left[\sqrt{\lambda} = 5\right]$ so $\lambda = 25$ | A1 | |
| $x = \frac{25}{2} \times 50$ so $x = 625$ m | dM1, A1 | (8) |
| | [16] | |
**Notes:**
| Item | Guidance |
|---|---|
| (a) | M1: Writing $P(X \leq 5) - P(X \leq 4)$ or $\frac{e^{-\lambda}\lambda^5}{5!}$ (any value of $\lambda$) | A1: awrt 0.0361 |
| (b) | 1st B1: Writing or using Po(8) | 2nd B1: awrt 0.313 (calc gives 0.3133742...) |
| | 1st M1: Recognize Binomial | 1st A1ft: writing $B(4, \text{'their } 0.313')$ May be $\Rightarrow$ by next line |
| | 2nd dM1 (dep. on 1st M1): $C_1^4 \times \text{'their } 0.3134 \times (1-\text{'their } 0.3134)^3$ | 2nd A1: awrt 0.406 |
| (c) | 1st M1: continuity correction used. Either 25.5 or 26.5 | 2nd M1: standardising using their $\lambda$ and $\sqrt{\lambda}$ for mean and sd. Any letter may be used or $\frac{\lambda}{25}$ etc |
| | 1st A1: $\frac{25.5 - \lambda}{\sqrt{\lambda}} = z$ where $0 < z < 0.5$. May be implied by their correct quadratic (25.5 req'd) |
| | B1: 0.1 (calc 0.09992...) used as their z value in an equation. Allow e.g. $\frac{26-\mu}{\sigma} = 0.1$ |
| | 3rd dM1 (dep on 2nd M1): some attempt at solving their 3TQ $= \frac{-b \pm \sqrt{b^2+4ac}}{2a}$ | 2nd A1: 25 (o.e.) |
| | 4th dM1 (dep on 3rd M1): $\text{their } \frac{25}{2} \times 50$ (If using $\frac{\lambda}{25}$ award when $x = $ ...) | 3rd A1: awrt 625 |7. Flaws occur at random in a particular type of material at a mean rate of 2 per 50 m .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that in a randomly chosen 50 m length of this material there will be exactly 5 flaws.
This material is sold in rolls of length 200 m . Susie buys 4 rolls of this material.
\item Find the probability that only one of these rolls will have fewer than 7 flaws.
A piece of this material of length $x \mathrm {~m}$ is produced.
Using a normal approximation, the probability that this piece of material contains fewer than 26 flaws is 0.5398
\item Find the value of $x$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2014 Q7 [16]}}