## Question 6:

### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X^N) = \int_0^{2\theta} \frac{x^{N+1}}{2\theta^2}dx$ | M1 | Attempting to integrate $\frac{x^{N+1}}{2\theta^2}$; $x^{N+1} \to x^{N+2}$; condone missing limits |
| $= \left[\frac{x^{N+2}}{2(N+2)\theta^2}\right]_0^{2\theta}$ | A1 | Correct integration |
| $= \frac{(2\theta)^{N+2}}{2(N+2)\theta^2} = \frac{2^{N+1}}{N+2}\theta^N$ | A1cso (3) | Must see substitution of $2\theta$ |

### Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = \frac{4\theta}{3}$ | B1 | Must have $E(X) = $; allow their $E(X)$ if one has been given otherwise must be correct |
| $\text{Var}(X) = 2\theta^2 - \left(\frac{\text{"4}\theta\text{"} }{3}\right)^2 = \frac{2\theta^2}{9}$ | M1A1 (3) | A1 must be using part (a); do not allow if integrated from scratch |

### Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $q = \frac{3}{4}$ | B1 | |
| $\text{Var}(S_1) = \frac{9}{16} \times \frac{\text{"}\frac{2\theta^2}{9}\text{"}}{n} = \frac{\theta^2}{8n}$ | M1, A1cso | M1 for $\frac{9}{16}\times\frac{\text{their Var}(X)}{n}$ |
| as $n\to\infty$, $\text{Var}(S) \to 0$; since it is unbiased, it is a consistent estimator | (3) | A1 cso and for $n\to\infty$, $\text{Var}(S)\to 0$; since it is unbiased it is a consistent estimator |

### Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(S_2) = a\times\frac{\text{"4}\theta\text{"}}{3} + b\times\frac{\theta}{3}$ | M1 | M1 for $a\times\text{their }E(X) + b\times\frac{\theta}{3}$ |
| $a\times\frac{4\theta}{3} + b\times\frac{\theta}{3} = \theta$ or $4a + b = 3$ | A1 | Correct equation with no $\theta$ |
| $\text{Var}(S_2) = a^2\times\frac{\text{"}\frac{2\theta^2}{9}\text{"}}{} + b^2\times\frac{\theta^2}{27}$ | M1 | M1 $a^2\times\text{their Var}(X) + b^2\times\frac{\theta^2}{27}$ |
| $\text{Var}(S_2) = a^2\times\frac{2\theta^2}{9} + (3-4a)^2\times\frac{\theta^2}{27}$ or $\text{Var}(S_2) = \left(\frac{3-b}{4}\right)^2\times\frac{2\theta^2}{9} + b^2\times\frac{\theta^2}{27}$ | M1 | M1 subst in for $a$ or $b$ |
| $\frac{d\text{Var}(S_2)}{da} = \frac{4a\theta^2}{9} - \frac{8(3-4a)\theta^2}{27} = 0$ or $\frac{-(3-b)\theta^2}{36} + \frac{2b\theta^2}{27} = 0$ | M1 | M1 differentiating with respect to $a$ or $b$ |
| $\frac{44a}{27} = \frac{24}{27}$ or $\frac{11}{108}b = \frac{1}{12}$ | M1 | M1 putting $d\text{Var}/da = 0$ and solving |
| $a = \frac{6}{11}$, $b = \frac{9}{11}$ | A1 (7) | Allow awrt 0.545 and awrt 0.818 |

### Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Var}(S_2) = \left(\frac{\text{"6"}}{11}\right)^2 \times \frac{2\theta^2}{9} + \left(\frac{\text{"9"}}{11}\right)^2 \times \frac{\theta^2}{27}$ | M1 | M1 subst $a$ and $b$ in to find $\text{Var}(S_2)$ |
| $\text{Var}(S_2) = \frac{\theta^2}{11}$ | (1) | |

### Part (f)
| Answer | Mark | Guidance |
|--------|------|----------|
| $S_1$ is the better estimator when $\frac{\theta^2}{8n} < \frac{\theta^2}{11} \Rightarrow n > \frac{11}{8}$ | M1 | M1 for reason $\frac{\theta^2}{8n} < \frac{\theta^2}{11} \Rightarrow n > \frac{11}{8}$ or $\text{Var}(S_1) \leq \frac{\theta^2}{16} < \frac{\theta^2}{11}$ |
| $S_2$ is the better estimator when $n < \frac{11}{8}$ | | |
| Therefore $S_1$ is the better estimator since $n \geq 2$ | A1cso (2) | Correct selection |

**Total: 19**

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