Pre-U Pre-U 9795/2 2018 June — Question 4

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2018
SessionJune
TopicCumulative distribution functions
TypePDF of transformed variable
DifficultyChallenging +1.2 This is a standard Further Maths statistics question requiring differentiation of a CDF to find the PDF, then calculating E(X) by integration, and applying the transformation formula for Y=1/X². While it involves multiple steps and the transformation technique, these are well-practiced procedures in Further Maths with no novel insight required. The transformation is straightforward once the formula is recalled.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03g Cdf of transformed variables

4 The continuous random variable \(X\) has cumulative distribution function given by $$\mathrm { F } ( x ) = \begin{cases} 0 & x < 0 \\ \frac { 1 } { 8 } x ^ { 3 } & 0 \leqslant x \leqslant 2 \\ 1 & x > 2 \end{cases}$$
  1. Find \(\mathrm { E } ( X )\).
  2. Find the probability density function of \(Y\), where \(Y = \frac { 1 } { X ^ { 2 } }\).

4 The continuous random variable $X$ has cumulative distribution function given by

$$\mathrm { F } ( x ) = \begin{cases} 0 & x < 0 \\ \frac { 1 } { 8 } x ^ { 3 } & 0 \leqslant x \leqslant 2 \\ 1 & x > 2 \end{cases}$$

(i) Find $\mathrm { E } ( X )$.\\
(ii) Find the probability density function of $Y$, where $Y = \frac { 1 } { X ^ { 2 } }$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2018 Q4}}