# Question 4:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(R < r) = k\pi r^2$ | M1 | Formulate probability proportional to area in terms of $k$ |
| $P(R < a) = 1 \therefore k = \dfrac{1}{\pi a^2}$ | M1 | Find $k$ |
| $\therefore P(R < r) = \dfrac{\pi r^2}{\pi a^2} = \dfrac{r^2}{a^2}$ | | IF M0M0, allow SC B1 for $P(R<r) = \dfrac{\pi r^2}{\pi a^2} = \ldots$ |
| Thus $F(r) = \dfrac{r^2}{a^2}$ (for $0 \le r \le a$). | A1 | Convincingly shown; ANSWER GIVEN. Condone omission of $r<0$ and/or $r>a$ |
| **[3]** | | |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| For $0 \le r \le a$, $f(r) = \dfrac{\mathrm{d}}{\mathrm{d}r}F(r) = \dfrac{2r}{a^2}$ | M1, A1 | Condone omission of $r<0$ and/or $r>a$ |
| **[2]** | | |

## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(R) = \int_0^a r \cdot \dfrac{2r}{a^2}\,\mathrm{d}r$ | M1 | Correct integral with limits (may be implied subsequently) |
| $= \left[\dfrac{2r^3}{3a^2}\right]_0^a$ | A1 | Correctly integrated |
| $= \dfrac{2a}{3}$ | A1 | Limits used. Accept unsimplified form |
| $E(R^2) = \int_0^a r^2 \cdot \dfrac{2r}{a^2}\,\mathrm{d}r = \left[\dfrac{2r^4}{4a^2}\right]_0^a = \dfrac{a^2}{2}$ | M1, A1 | Correct integral with limits; correctly integrated and limits used |
| $\text{Var}(R) = \dfrac{a^2}{2} - \left(\dfrac{2a}{3}\right)^2 = \dfrac{9a^2 - 8a^2}{18} = \dfrac{a^2}{18}$ | M1, A1 | Use of $\text{Var}(R) = E(R^2) - [E(R)]^2$; Convincingly shown — ANSWER GIVEN. Require sight of both terms with common denominator |
| **[7]** | | |

## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{R} \sim (\text{approx})\ N\!\left(\dfrac{2}{3}\times 22.5,\ \dfrac{22.5^2}{18\times 100}\right) = N\!\left(15,\ 0.28125\right)$ | B1, B1, B1 | Normal; Mean — ft c's $E(R)$ (>0) with $a=22.5$; Variance — cao ($= 0.5303(3)^2$). Accept unsimplified form |
| **[3]** | | |

## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER**: can argue that 13.87 is more than 2 SDs from the Mean (15). $15 - 2\sqrt{0.28125} = 13.93(9)$, must refer to $\text{SD}(\bar{R})$, not $\text{SD}(R)$, i.e. outlier $\Rightarrow$ doubt. | M1, M1, A1 | Allow 1.96 SDs, but not 1.984. 1.96 gives 13.96(1). A 95% CI is (12.831, 14.909). Must see explicit evidence. ft c's mean |
| **OR** more formally like a significance test: refer to $N(0,1)$, $\dfrac{13.87-15}{\sqrt{0.28125}} = -2.131$, sig at (eg) 5% $\Rightarrow$ doubt. | M1, M1, A1 | Could imply first M. $P(\|Z\| > 2.131) = 0.0332$. ft c's mean |
| **[3]** | | |

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