# Question 7:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim U[0, 9]$ | B1 | For $X \sim U[0,9]$ or "continuous uniform"/"rectangular" distribution with correct range. Or allow pdf $f(x) = \frac{1}{9}$, $0 \leq x \leq 9$, $0$ otherwise |
| | **(1)** | |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 6) = \frac{1}{3}$ oe, allow awrt 0.333 | B1 | |
| | **(1)** | |

## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = X(9-X)$, $= 9X - X^2$ | M1, A1 | M1 for $X(9-X)$ or $9X - X^2$ may be implied by correct answer. A1 for $9X - X^2$ or $a=-1$ and $b=9$ |
| | **(2)** | |

## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = 4.5$ | B1 | 1st B1 for 4.5, may be implied |
| $\text{Var}(X) = \frac{81}{12} = \frac{27}{4}$ **or** $E(X^2) = \int_0^9 \frac{x^2}{9}\, dx$ | B1 | 2nd B1 for $\frac{81}{12}$ or $\frac{27}{4}$ **or** $\int_0^9 \frac{x^2}{9}$, ignore limits |
| $E(X^2) = \text{Var}(X) + [E(X)]^2$ **or** $= \left[\frac{x^3}{27}\right]_0^9$ | M1 | 1st M1 for full method for $E(X^2)$ using $\text{Var}(X)$ and $E(X)$, or attempt to integrate $x^n \to x^{n+1}$ leading to a value for $E(X^2)$. Need to be using $\int_0^9 \frac{x^2}{9}$, ignore limits |
| $E(X^2) = 27$ | A1 | 1st A1 for $E(X^2) = 27$, may be implied |
| $E(R) = 9 \times 4.5 - 27 = 13.5$ | dM1A1 | d2nd M1 for using $9E(X) - E(X^2)$, with their $E(X)$ and $E(X^2)$. Dep on first M |
| | **(6)** | |

**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_0^9 \frac{(9x - x^2)}{9}\, dx = \left[\frac{9x^2}{18} - \frac{x^3}{27}\right]_0^9$ | B1 B1 M1A1 | B1 for integrand (ignore limits), ft their (c) which must be of form $aX^2 + b$. B1 with correct limits ft their (c). M1 attempt to integrate at least one $x^n \to x^{n+1}$. A1 correct integration |
| $= \frac{81}{2} - \frac{81}{3} = 13.5$ | dM1, A1 | dM1 subst in limits, need to see 9 substituted. Condone missing 0 |

## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $R > 2X^2$ or $9X - X^2 > 2X^2$ | M1 | 1st M1 for forming suitable inequality in $R$ and $X$ or just $X$. May be implied by a correct probability in $X$. Allow $\leq$ instead of $<$ and $\geq$ instead of $>$ |
| $9X > 3X^2$ | A1 | 1st A1 for simplifying to $9X > 3X^2$ or $3 > X$. May be implied by correct probability in $X$ |
| So $P(X < 3)$ | M1 | 2nd M1 for forming a correct probability in $X$ |
| $= \frac{1}{3}$ | A1 | 2nd A1 for $\frac{1}{3}$ or exact equivalent |
| | **(4)** | |
| | **Total: (14)** | |