# Question 7:

## Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int_{-3}^{0} c(x+3)\,dx + \int_{0}^{3} \frac{5}{36}(3-x)\,dx = 1$ **or** area of triangle $= \frac{3\times 3c}{2} + \frac{3\times \frac{5}{36}(3)}{2}$ | M1 | Correct integrals added with limits, or area of triangle. Do not accept $c(0+3)=\frac{5}{36}(3-0)$ |
| $c\left(-\frac{9}{2}+9\right)+\frac{5}{8}=1$ oe **or** $\frac{3\times 3c}{2}+\frac{5}{8}=1$ oe | dM1 | Equating to 1 and attempting to solve |
| $c = \frac{1}{12}$ * | A1cso | No incorrect working seen |

**Alternative:**
| $\left[\int c(x+3)\right] = c\left(\frac{x^2}{2}+3x\right)+d$ and $\left[\int\frac{5}{36}(3-x)\right]=\frac{5}{36}\left(3x-\frac{x^2}{2}\right)+e$ | M1 | |
| Using $F(-3)=0$ and $F(3)=1$ and $d=e$ | dM1 | |
| $c=\frac{1}{12}$ * | A1 cso | |

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## Part (b)(i):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct sketch of $f(x)$: straight line positive gradient from $-3\leq x<0$ and straight line negative gradient from $0\leq x\leq 3$, LHS below RHS at $y$-axis | B1 | |
| Correct labels: $-3$ and $3$ on $x$-axis and $\frac{1}{4}$ and $\frac{5}{12}$ oe on $y$-axis | B1 | |

## Part (b)(ii):

| Working/Answer | Mark | Guidance |
|---|---|---|
| **Highest point** [at $f(0)$, so $X=0$ is the mode] oe | B1 | oe e.g. Maximum value. Condone "highest probability" |

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## Part (c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int_{-3}^{x}\frac{1}{12}(t+3)\,dt$ **or** $\int\frac{1}{12}(x+3)\,dx$ with $+c$ and $F(-3)=0$ | M1 | Also: attempting to integrate second line with correct limits $+F(0)$ **or** $+d$ and $F(3)=1$. Allow use of $F(0)=\frac{3}{8}$ instead of $F(-3)=0$ or $F(3)=1$ |
| $F(x)=\begin{cases}0 & x<-3\\ \frac{1}{24}x^2+\frac{1}{4}x+\frac{3}{8} & -3\leq x<0\\ \frac{5}{12}x-\frac{5}{72}x^2+\frac{3}{8} & 0\leq x\leq 3\\ 1 & x>3\end{cases}$ **or** $F(x)=\begin{cases}0\\ \frac{1}{24}(x+3)^2\\ 1-\frac{5}{72}(3-x)^2\\ 1\end{cases}$ | B1, A1, A1 | B1: 1st and 4th lines correct (allow $\leq$ instead of $<$ and vice versa). A1: 2nd line correct with limits. A1: 3rd line correct with limits |

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## Part (d):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $[P(X>d\mid X>0)=]\;\frac{1-F(d)}{1-F(0)}=\frac{2}{5}$ oe | M1 | Correct probability expression. Allow $\frac{P(X>d)}{P(X>0)}=\frac{2}{5}$. Do not allow $P(X>d)\cap P(X>0)$ |
| $1-F(d)=\frac{2}{5}\times\frac{5}{8}\left[=\frac{1}{4}\right]$ oe | A1 | Or $1-P(x<d)=\frac{2}{5}\times\frac{5}{8}=\frac{1}{4}$, or $F(d)=\frac{3}{4}$, or $P(x>d)=\frac{1}{4}$ |
| $1-\left(\frac{5}{12}d-\frac{5}{72}d^2+\frac{3}{8}\right)=\frac{2}{5}\times\frac{5}{8}$ **or** $\frac{(3-d)\frac{5}{36}(3-d)}{2}=\frac{2}{5}\times\frac{5}{8}$ oe | dM1 | Previous M1 awarded. Using their cdf or area of triangle, setting $1-F(d)=$ their $P(X>d)$ and solving |
| $d=$ awrt $\mathbf{1.10}$ | A1 | |