| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with binomial combination |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with standard techniques: basic probability calculations, scaling parameters, combining independent Poisson variables, and applying normal approximation to binomial. All parts follow textbook methods with no novel problem-solving required, though the multi-part structure and distribution combination in (iv) elevate it slightly above average difficulty. |
| 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 probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 'Independently' means that the occurrence of one defect does not affect the probability of another defect occurring | E1 | Allow e.g. "event" for "defect" |
| Poisson distribution | E1 | Allow Po(...) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(2 \text{ defects}) = \frac{e^{-1.7}1.7^2}{2!}\) OR from tables \(P(2 \text{ defects}) = 0.7572 - 0.4932\) | M1 | For calculation of \(P(X=2)\) |
| \(= 0.2640\) | A1 | CAO; allow 0.264; Do not allow 0.2639 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| New \(\lambda = 7 \times 1.7 = 11.9\) | B1 | For mean (SOI) |
| \(P(12 \text{ defects}) = \frac{e^{-11.9}11.9^{12}}{12!}\) | M1 | For calculation with their \(\lambda \neq 1.7\) |
| \(= 0.1143\) cao | A1 | Allow 0.114 www; Note B0M1A0 for 0.114 from use of \(\lambda = 11.7\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| New \(\lambda = (2 \times 1.7) + (2 \times 0.8) = 5.0\) | B1* | For new mean \(= 5\) seen |
| \(P(\text{at least 8 defects}) = 1 - P(7 \text{ or fewer defects})\) | B1* | For \(1 - P(X \leq 7)\) or \(1 - P(X < 8)\) seen |
| \(= 1 - 0.8666\) | B1dep* | For \(1 - 0.8666\) seen |
| \((= 0.1334)\) NB Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Binomial\((100, 0.1334)\) or \(B(100, 0.1334)\) | B1* | For binomial |
| B1dep* | For parameters |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 13.34\) | B1 | For mean (soi) |
| Variance \(= 100 \times 0.1334 \times 0.8666 = 11.56\) | B1 | For variance (soi); For \(B(13.34, 11.56)\) seen, award B0B0 unless used correctly as part of a Normal calculation |
| Using Normal approx. to the binomial, \(X \sim N(13.34, 11.56)\) | ||
| \(P(X \geq 20) = P\left(Z \geq \frac{19.5 - 13.34}{\sqrt{11.56}}\right)\) | B1 | For 19.5 seen |
| M1 | For probability using correct tail and a sensible calculation with their mean and variance; e.g. using standard deviation \(= 11.56\) or finding \(P(Z < 1.812)\) gets M0A0 | |
| \(= P(Z > 1.812) = 1 - \Phi(1.812) = 1 - 0.9650\) | ||
| \(= 0.035\) | A1 | CAO (Do not FT wrong or omitted CC) |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 'Independently' means that the occurrence of one defect does not affect the **probability** of another defect occurring | E1 | Allow e.g. "event" for "defect" |
| Poisson distribution | E1 | Allow Po(...) |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(2 \text{ defects}) = \frac{e^{-1.7}1.7^2}{2!}$ **OR** from tables $P(2 \text{ defects}) = 0.7572 - 0.4932$ | M1 | For calculation of $P(X=2)$ |
| $= 0.2640$ | A1 | CAO; allow 0.264; Do not allow 0.2639 |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| New $\lambda = 7 \times 1.7 = 11.9$ | B1 | For mean (SOI) |
| $P(12 \text{ defects}) = \frac{e^{-11.9}11.9^{12}}{12!}$ | M1 | For calculation with their $\lambda \neq 1.7$ |
| $= 0.1143$ cao | A1 | Allow 0.114 www; Note B0M1A0 for 0.114 from use of $\lambda = 11.7$ |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| New $\lambda = (2 \times 1.7) + (2 \times 0.8) = 5.0$ | B1* | For new mean $= 5$ seen |
| $P(\text{at least 8 defects}) = 1 - P(7 \text{ or fewer defects})$ | B1* | For $1 - P(X \leq 7)$ or $1 - P(X < 8)$ seen |
| $= 1 - 0.8666$ | B1dep* | For $1 - 0.8666$ seen |
| $(= 0.1334)$ **NB Answer given** | | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Binomial$(100, 0.1334)$ or $B(100, 0.1334)$ | B1* | For binomial |
| | B1dep* | For parameters |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 13.34$ | B1 | For mean (soi) |
| Variance $= 100 \times 0.1334 \times 0.8666 = 11.56$ | B1 | For variance (soi); For $B(13.34, 11.56)$ seen, award B0B0 unless used correctly as part of a Normal calculation |
| Using Normal approx. to the binomial, $X \sim N(13.34, 11.56)$ | | |
| $P(X \geq 20) = P\left(Z \geq \frac{19.5 - 13.34}{\sqrt{11.56}}\right)$ | B1 | For 19.5 seen |
| | M1 | For probability using correct tail and a sensible calculation with their mean and variance; e.g. using standard deviation $= 11.56$ or finding $P(Z < 1.812)$ gets M0A0 |
| $= P(Z > 1.812) = 1 - \Phi(1.812) = 1 - 0.9650$ | | |
| $= 0.035$ | A1 | CAO (Do not FT wrong or omitted CC) |
---2 Manufacturing defects occur in a particular type of aluminium sheeting randomly, independently and at a constant average rate of 1.7 defects per square metre.\\
\begin{enumerate}[label=(\roman*)]
\item Explain the meaning of the term 'independently' and name the distribution that models this situation.
\item Find the probability that there are exactly 2 defects in a sheet of area 1 square metre.
\item Find the probability that there are exactly 12 defects in a sheet of area 7 square metres.
In another type of aluminium sheet, defects occur randomly, independently and at a constant average rate of 0.8 defects per square metre.
\item A large box is made from 2 square metres of the first type of sheet and 2 square metres of the second type of sheet, chosen independently. Show that the probability that there are at least 8 defects altogether in the box is 0.1334 .
A random sample of 100 of these boxes is selected.
\item State the exact distribution of the number of boxes which have at least 8 defects.
\item Use a suitable approximating distribution to find the probability that there are at least 20 boxes in the sample which have at least 8 defects.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2014 Q2 [17]}}