# Question 2:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(w) = \begin{cases} \frac{1}{8} & -1.4 < w < 6.6 \\ 0 & \text{otherwise} \end{cases}$ | M1 | pdf of this form where $p$ is a probability; allow $\leq$ instead of $<$; allow equivalent for the 0 otherwise |
| | A1 | Fully correct; allow $\leq$ instead of $<$. Allow any letter but must be consistent |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(W) = 2.6$ oe | B1 | 2.6 oe |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1.6 - \alpha) \times \frac{1}{8} = 0.35$ | M1 | Setting up equation $(1.6-\alpha)\times$ "their $p$" $= 0.35$ with $0<p<1$, or $\frac{7}{20}=\frac{2.8}{8}$ and $\alpha = 1.6-$"2.8"; or $F(1.6)-F(\alpha)=0.35$ using F(w) in form $bw+c$ where $0<b<1$; allow $\int_\alpha^{1.6}$"their f(w)"dw $=0.35$ with attempt to integrate |
| $\alpha = -1.2$ oe | A1cso | If using $F(1.6)-F(\alpha)=0.35$ then F(w) must be correct |

## Part (d)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(1.2 < W < 2.4) = (2.4-1.2)\times\frac{1}{8}$ | M1 | $(2.4-1.2)\times$ "their $p$" where $\frac{1}{8}$ is a probability, or $F(2.4)-F(1.2)$ using F(w) in form $bw+c$, $0<b<1$. Implied by 0.15 |
| $= \frac{3}{20}$ or 0.15 oe | A1ft | Ft their $p$ as long as answer is a probability |

## Part (e)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(W>2 \mid 1.2<W<2.4) = \frac{0.4\times\frac{1}{8}}{\text{"0.15"}}$ | M1 | $\frac{0.4\times\text{"their }\frac{1}{8}\text{"}}{\text{"their (d)"}}$ or $\frac{0.4}{\text{"1.2"}}$ implied by $\frac{1}{3}$; allow $\int_2^{2.4}$"their f(w)"dw for numerator |
| $= \frac{1}{3}$ (awrt 0.333) | A1 | Allow $0.\dot{3}$ or $0.\overline{3}$ |

## Part (f)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y \sim B(40, \text{"0.15"})$ | M1 | Writing or using $B(40,$ "their 0.15"). Implied by mean of $40\times$"their (d)" |
| $P(Y\geq10) = 1-P(Y\leq9)$ or $1-0.9328$ | M1 | Writing or using $1-P(Y\leq 9)$. Allow for $1-P\!\left(z\leq\frac{9.5 \text{ or } 9 - \text{"their mean"}}{\text{"their sd"}}\right)$ |
| $= 0.0672$ (awrt 0.0672) | A1 | awrt 0.0672 |