# Question 7:

## Part (a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \sum_{x=0}^{\infty} x \cdot \frac{e^{-\lambda}\lambda^x}{x!}$ | M1 | Must start from definition of $E(X)$ |
| $= \sum_{x=1}^{\infty} \frac{e^{-\lambda}\lambda^x}{(x-1)!}$ | M1 | Cancel $x$ with $x!$, lower limit changes to 1 |
| $= \lambda e^{-\lambda} \sum_{x=1}^{\infty} \frac{\lambda^{x-1}}{(x-1)!} = \lambda e^{-\lambda} \cdot e^{\lambda} = \lambda$ | A1 | Recognition of exponential series summing to $e^{\lambda}$ |

## Part (a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(X) = E(X^2) - [E(X)]^2 = E(X^2 - X) + E(X) - [E(X)]^2 = \lambda^2 + \lambda - \lambda^2 = \lambda$ | B1 | Must use given result and part (i) |

## Part (b)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 10$, $H_1: \lambda > 10$ | B1 | Both hypotheses correct |
| $X \sim \text{Po}(10)$, find $P(X \geq 15)$ | M1 | Correct tail |
| $P(X \geq 15) = 1 - P(X \leq 14) = 1 - 0.9165 = 0.0835$ | A1 | Correct probability |
| $0.0835 > 0.05$, do not reject $H_0$ | M1 | Correct comparison |
| Insufficient evidence that $\lambda > 10$ | A1 | Correct conclusion in context |

## Part (b)(ii)(A):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Total faults $\sim \text{Po}(20\lambda)$; under $H_0$: $\text{Po}(200)$ | B1 | Correct distribution stated |
| Approximate with $N(200, 200)$ | M1 | Normal approximation |
| $z = \frac{241 - 0.5 - 200}{\sqrt{200}} = \frac{240.5 - 200}{14.142} = 2.864$ | M1 | Continuity correction and correct standardisation |
| $2.864 > 1.6449$, reject $H_0$ | A1 | Correct conclusion — sufficient evidence $\lambda > 10$ |

## Part (b)(ii)(B):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Find $c$ such that $P(X \geq c) \leq 0.05$ where $X \sim N(200, 200)$ | M1 | Correct method |
| $\frac{c - 0.5 - 200}{\sqrt{200}} = 1.6449$ | M1 | Using $z = 1.6449$ with continuity correction |
| $c = 200 + 0.5 + 1.6449\sqrt{200} = 224.26...$ | A1 | Critical value = **225** |

## Part (b)(ii)(C):

| Answer/Working | Mark | Guidance |
|---|---|---|
| If $\lambda = 12$, total $\sim \text{Po}(240) \approx N(240, 240)$ | B1 | Correct distribution |
| Type II error: $P(\text{not reject } H_0 \mid \lambda = 12) = P(X < 225)$ | M1 | Correct definition of Type II error applied |
| $P\left(Z < \frac{224.5 - 240}{\sqrt{240}}\right) = P(Z < -1.000)$ | M1 | Correct standardisation with continuity correction |
| $= 1 - \Phi(1.000) = 1 - 0.8413 = 0.1587$ | A1 | Correct probability |

These pages (22, 23, and 24) from the AQA June 2014 paper (P71758/Jun14/MS03) contain no mark scheme content. They are all blank "continuation" pages with only the text:

- "There are no questions printed on this page"
- "DO NOT WRITE ON THIS PAGE"
- "ANSWER IN THE SPACES PROVIDED"

There is no mark scheme content to extract from these pages.