| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with binomial combination |
| Difficulty | Standard +0.3 This is a straightforward S2 question combining standard Poisson calculations with binomial probability. Part (a) is direct Poisson lookup, part (b) applies binomial with the result from (a), and part (c) uses normal approximation to Poisson—all routine techniques for this module with no novel problem-solving required. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02d Binomial: mean np and variance np(1-p)5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04b Linear combinations: of normal distributions5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim Po(5)\); \(P(X \leq 3) = 0.2650\) | M1 A1 | M1 for identifying \(Po(5)\) — should be clearly seen or implied; A1 for correct probability, allow 0.265 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(Y\) = no. of planks with at most 3 defects, \(Y \sim B(6, 0.265)\) | M1 A1ft | \(1^\text{st}\) M1 for writing or using binomial; A1ft for \(n=6\), \(p=\) their (a) |
| \(P(Y<2) = P(Y \leq 1) = \left[0.735^6 + 6 \times 0.265 \times 0.735^5\right]\) | M1 A1 | \(2^\text{nd}\) M1 for \(P(Y \leq 1)\) or \(P(Y=0)+P(Y=1)\) or \((1-q)^6 + nq(1-q)^5\) |
| \(= 0.4987\ldots\) awrt 0.499 or 0.498 | A1 | \(3^\text{rd}\) A1 for awrt 0.499; SC use of tables — lose last two marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(T\) = total number of defects on 6 planks, \(T \sim Po(30)\) so \(T \approx S \sim \text{Normal}\) | M1 | \(1^\text{st}\) M1 for normal approximation |
| \(S \sim N(30, 30)\) | A1 | \(1^\text{st}\) A1 for correct mean and sd |
| \(P(T < 18) = P(S < 17.5)\) | M1 | \(2^\text{nd}\) M1 for continuity correction, either 17.5 or 18.5 or 42.5 or 41.5 seen |
| \(= P\!\left(z < \frac{17.5 - 30}{\sqrt{30}}\right)\) | M1 | \(3^\text{rd}\) M1 standardising with mean, sd and 17.5 or 18 or 18.5 or 41.5 or 42 or 42.5 |
| \(= P(Z < -2.28\ldots)\) | A1 | \(2^\text{nd}\) A1 for \(z = \pm 2.28\) or better |
| \(= 0.01123\ldots\) awrt 0.0112 or 0.0113 | A1 | \(3^\text{rd}\) A1 for awrt 0.0112 or 0.0113; NB no approximation gives 0.00727 |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim Po(5)$; $P(X \leq 3) = 0.2650$ | M1 A1 | M1 for identifying $Po(5)$ — should be clearly seen or implied; A1 for correct probability, allow 0.265 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $Y$ = no. of planks with at most 3 defects, $Y \sim B(6, 0.265)$ | M1 A1ft | $1^\text{st}$ M1 for writing or using binomial; A1ft for $n=6$, $p=$ their (a) |
| $P(Y<2) = P(Y \leq 1) = \left[0.735^6 + 6 \times 0.265 \times 0.735^5\right]$ | M1 A1 | $2^\text{nd}$ M1 for $P(Y \leq 1)$ or $P(Y=0)+P(Y=1)$ or $(1-q)^6 + nq(1-q)^5$ |
| $= 0.4987\ldots$ awrt 0.499 or 0.498 | A1 | $3^\text{rd}$ A1 for awrt 0.499; SC use of tables — lose last two marks |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $T$ = total number of defects on 6 planks, $T \sim Po(30)$ so $T \approx S \sim \text{Normal}$ | M1 | $1^\text{st}$ M1 for normal approximation |
| $S \sim N(30, 30)$ | A1 | $1^\text{st}$ A1 for correct mean and sd |
| $P(T < 18) = P(S < 17.5)$ | M1 | $2^\text{nd}$ M1 for continuity correction, either 17.5 or 18.5 or 42.5 or 41.5 seen |
| $= P\!\left(z < \frac{17.5 - 30}{\sqrt{30}}\right)$ | M1 | $3^\text{rd}$ M1 standardising with mean, sd and 17.5 or 18 or 18.5 or 41.5 or 42 or 42.5 |
| $= P(Z < -2.28\ldots)$ | A1 | $2^\text{nd}$ A1 for $z = \pm 2.28$ or better |
| $= 0.01123\ldots$ awrt 0.0112 or 0.0113 | A1 | $3^\text{rd}$ A1 for awrt 0.0112 or 0.0113; NB no approximation gives 0.00727 |
**SC** using $P(X<18.5) - P(X<17.5)$ can get M1 A1 M1 M0A0A0
---5. Defects occur at random in planks of wood with a constant rate of 0.5 per 10 cm length. Jim buys a plank of length 100 cm .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that Jim's plank contains at most 3 defects.
Shivani buys 6 planks each of length 100 cm .
\item Find the probability that fewer than 2 of Shivani's planks contain at most 3 defects.
\item Using a suitable approximation, estimate the probability that the total number of defects on Shivani's 6 planks is less than 18.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2011 Q5 [13]}}