# 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) |

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