| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Multiple periods with binomial structure |
| Difficulty | Standard +0.3 This is a straightforward multi-part Poisson question requiring: (a) reverse lookup in tables (routine), (b) normal approximation with continuity correction (standard S2 technique), and (c) binomial calculation using the probability from part (a). All parts follow standard textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d 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 probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X\) = faults in a week \(\Rightarrow X \sim \text{Po}(6)\) | — | — |
| \([P(X \geqslant x) = 0.1528 \Rightarrow P(X \leqslant x-1)] = 0.8472\) | M1 | Writing or using \(1 - P(X \leqslant x-1)\) |
| Using tables \(P(X \leqslant 8) = 0.8472 \Rightarrow x - 1 = 8\) | M1 | For 0.8472; may be implied by \(x - 1 = 8\) |
| \(x = 9\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Y\) = faults in six weeks \(\Rightarrow Y \sim \text{N}(36, 36)\) | B1 | Writing or using \(\text{N}(36,36)\); may be implied by a correct standardisation expression |
| \(P(Y < 32) = P\!\left(Z < \dfrac{31.5 - 36}{6}\right) \left[= P(Z < -0.75)\right]\) | M1 M1 | Standardising with 30.5/31/31.5/32/32.5/39.5/40/40.5/41/41.5 (Allow \(\pm\)); fully correct standardisation, may be implied by \(\pm 0.75\) |
| \(= 0.2266\) (awrt 0.227) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(W\) = Number of *poor weeks* \(\Rightarrow W \sim \text{B}(50, 0.1528)\) | B1 | Writing or using \(\text{B}(50, 0.1528)\) |
| \([P(W > 1)] = 1 - P(W \leqslant 1)\) | M1 | Writing or using \(1 - P(W \leqslant 1)\) (allow any letter) |
| \(= 1 - \!\left(0.8472^{50} + 50 \times 0.1528 \times 0.8472^{49}\right)\) | dM1 | Dependent on using binomial; using \(1-[P(W=0)+P(W=1)]\) (implied by awrt 0.997 or 0.9975 or \(1 -\) awrt 0.00257); using binomial may be implied by \((1-p)^{50} + {}^nC_r \times p \times (1-p)^{49}\) where \(p\) is a probability; condone \({}^nC_r\) missing |
| \(= 0.99748\ldots\) (awrt 0.997) | A1 | awrt 0.997 or 0.9975 |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X$ = faults in a week $\Rightarrow X \sim \text{Po}(6)$ | — | — |
| $[P(X \geqslant x) = 0.1528 \Rightarrow P(X \leqslant x-1)] = 0.8472$ | M1 | Writing or using $1 - P(X \leqslant x-1)$ |
| Using tables $P(X \leqslant 8) = 0.8472 \Rightarrow x - 1 = 8$ | M1 | For 0.8472; may be implied by $x - 1 = 8$ |
| $x = 9$ | A1 | — |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y$ = faults in six weeks $\Rightarrow Y \sim \text{N}(36, 36)$ | B1 | Writing or using $\text{N}(36,36)$; may be implied by a correct standardisation expression |
| $P(Y < 32) = P\!\left(Z < \dfrac{31.5 - 36}{6}\right) \left[= P(Z < -0.75)\right]$ | M1 M1 | Standardising with 30.5/31/31.5/32/32.5/39.5/40/40.5/41/41.5 (Allow $\pm$); fully correct standardisation, may be implied by $\pm 0.75$ |
| $= 0.2266$ (awrt 0.227) | A1 | — |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $W$ = Number of *poor weeks* $\Rightarrow W \sim \text{B}(50, 0.1528)$ | B1 | Writing or using $\text{B}(50, 0.1528)$ |
| $[P(W > 1)] = 1 - P(W \leqslant 1)$ | M1 | Writing or using $1 - P(W \leqslant 1)$ (allow any letter) |
| $= 1 - \!\left(0.8472^{50} + 50 \times 0.1528 \times 0.8472^{49}\right)$ | dM1 | Dependent on using binomial; using $1-[P(W=0)+P(W=1)]$ (implied by awrt 0.997 or 0.9975 or $1 -$ awrt 0.00257); using binomial may be implied by $(1-p)^{50} + {}^nC_r \times p \times (1-p)^{49}$ where $p$ is a probability; condone ${}^nC_r$ missing |
| $= 0.99748\ldots$ (awrt 0.997) | A1 | awrt 0.997 or 0.9975 |
**Total: 11 marks**1 A local pottery makes cups. The number of faulty cups made by the pottery in a week follows a Poisson distribution with a mean of 6
In a randomly chosen week, the probability that there will be at least $x$ faulty cups made is 0.1528
\begin{enumerate}[label=(\alph*)]
\item Find the value of $x$
\item Use a normal approximation to find the probability that in 6 randomly chosen weeks the total number of faulty cups made is fewer than 32
A week is called a "poor week" if at least $x$ faulty cups are made, where $x$ is the value found in part (a).
\item Find the probability that in 50 randomly chosen weeks, more than 1 is a "poor week".
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\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2022 Q1 [11]}}