# Question 3:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X \neq 4) = 1 - P(X=4)$, $= 1 - \frac{e^{-7}7^4}{4!}$ or $1-(0.1730-0.0818)$ | M1 | For $1-P(X=4)$ or $1-P(X \leq 4)+P(X \leq 3)$ oe |
| $= 0.90877\ldots$ | A1 awrt 0.909 | |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(Y=1) = (1-\text{"0.90877..."})(\text{"0.90877..."})^4 \times {}^5C_1$ | M1M1 | $(1-\text{"their 0.909"})^1(\text{"their 0.909"})^4$ or $(1-\text{"their 0.909"})(\text{"their 0.909"})^4$ allow values to 2s.f.; $P(Y=1)=(1-\text{"their 0.909"})(\text{"their 0.909"})^4 \times {}^5C_1$ allow values to 2s.f. |
| $= 0.311\ldots$ | A1 | awrt 0.312 or awrt 0.311 |

## Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda = 0.07n$ | B1 | Writing or using mean as $0.07n$ |
| $A \sim N(0.07n, 0.07n)$ | M1 | Normal with mean = variance which must be in terms of $n$ |
| $\frac{3.5 - \text{"0.07n"}}{\sqrt{\text{"0.07n"}}}$ | M1 | Standardising with their mean and their $\sqrt{\text{var}}$; allow 2.5, 3, 3.5, 4, 4.5 (correct standardisation implies B1M1M1) |
| $\frac{3.5-0.07n}{\sqrt{0.07n}} = -1.55$ or $\text{"0.07n"} - (1.55\sqrt{0.07})\sqrt{n} - 3.5 = 0$ | B1 | Their standardisation $= \pm 1.55$ |
| $n - \left(\frac{1.55}{0.07}\sqrt{0.07}\right)\sqrt{n} - \frac{3.5}{0.07} = 0 \Rightarrow n - 1.55\sqrt{\frac{n}{0.07}} - 50 = 0$ | A1cso | Must come from compatible signs; need at least one step between standardisation indicating division by 0.07 and correct equation |

## Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{n} = \frac{\frac{1.55}{\sqrt{0.07}} \pm \sqrt{\left(\frac{1.55}{\sqrt{0.07}}\right)^2 + 4\times 50}}{2} = \text{awrt } {-4.72}\ldots \text{ or awrt } 10.6\ldots\ (4\sqrt{7})$ | M1 | Correct method to solve given quadratic or sight of awrt $-4.72$ or awrt $10.6$ |
| $n = 112$ | A1cao | 112 only (must reject 2nd answer if found); an answer of 112 only scores M1A1 |

## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 7 \quad H_1: \lambda > 7$ | B1 | Both hypotheses correct in terms of $\lambda$ or $\mu$ [using $p$ scores B0] |
| $P(X \geq 15) = 1 - P(X \leq 14)$, $P(X \geq 14) = 0.0128$ | M1 | For $1-P(X \leq 14)$ or for CR: one of $P(X \geq 14)=0.0128$ or $P(X \geq 15)=0.0057$ |
| $= 1 - 0.9943$, $P(X \geq 15) = 0.0057$ | |  |
| $= 0.0057$, CR $X \geq 15$ | A1 | awrt 0.0057 or correct CR allow $X > 14$ |
| Reject $H_0$, in the CR, Significant | dM1 | (dep on 1st M1) A correct non-contextual statement consistent with their prob and 0.01 |
| There is evidence that the number of water fleas per 100 ml of the pond water has **increased** | A1 | All previous marks must be awarded. Correct context, conclusion with increase(oe) and fleas |

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