| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Topic | Cumulative distribution functions |
| Type | PDF of transformed variable |
| Difficulty | Challenging +1.2 This is a structured multi-part question on transforming random variables with clear guidance. Part (i) is routine integration, part (ii) provides the key algebraic relationship to show, and the PDF transformation follows standard technique. Parts (iii) and (iv) require straightforward integration. While it involves several steps and the transformation of variables concept, the question scaffolds the solution heavily and uses standard methods throughout. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
(i) $F(x) = \begin{cases} 0 & x < 0 \\ \frac{x}{2} & 0 \leqslant x \leqslant 2 \\ 1 & x > 2 \end{cases}$ **B1B1** [2]
(ii) $Y = X(8 - X) = 8X - X^2 \Rightarrow 16 - Y = 16 - 8X + X^2 = (4-X)^2$ (AG) **M1A1**
$G(y) = \frac{1}{2}\{4 - \sqrt{16 - y}\}$, $0 \leqslant y \leqslant 12$ **B1B1**
$\therefore g(y) = G'(y) = \frac{1}{4\sqrt{16-y}}$, $0 \leqslant y \leqslant 12$ (AG) **B1** [5]
(iii) $G(\text{Median}) = \frac{1}{2} \Rightarrow \text{Median} = 7$ **M1A1** [2]
(iv) $E(Y) = \int_0^{12} \frac{y}{4\sqrt{16-y}}\,dy$ **B1**
Let $16 - y = u^2$ (or any other valid substitution) $E(Y) = \int_2^4 (8 - \frac{1}{2}u^2)\,du$ **M1A1**
$E(Y) = \left[8u - \frac{u^3}{6}\right]_2^4 = \left[32 - \frac{32}{3}\right] - \left[16 - \frac{4}{3}\right] = \frac{20}{3}$ **M1A1** [5]6 A rectangle of area $Y \mathrm {~m} ^ { 2 }$ has a perimeter of 16 m and a side of length $X \mathrm {~m}$, where $X$ is a random variable with probability density function, f, given by
$$f ( x ) = \begin{cases} \frac { 1 } { 2 } & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
(i) Obtain the cumulative distribution function, F , of $X$.\\
(ii) Show that
$$16 - Y = ( 4 - X ) ^ { 2 }$$
and deduce that the probability density function of the random variable $Y$ is
$$g ( y ) = \begin{cases} \frac { 1 } { 4 \sqrt { 16 - y } } & 0 \leqslant y \leqslant 12 \\ 0 & \text { otherwise } \end{cases}$$
(iii) Find the median of $Y$.\\
(iv) Find $\mathrm { E } ( Y )$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2013 Q6 [14]}}