CAIE FP2 2011 June — Question 11 OR

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeGeometric/graphical PDF with k
DifficultyStandard +0.8 This Further Pure question requires finding k from the normalization condition, deriving the CDF by integrating piecewise linear functions, then applying transformation of variables (Y=X²) requiring the Jacobian method and careful domain mapping. The median calculation involves solving a cubic equation. Multiple sophisticated techniques beyond standard A-level, but systematic application of learned methods.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables

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The continuous random variable \(X\) takes values in the interval \(0 \leqslant x \leqslant 3\) only. For \(0 \leqslant x \leqslant 3\) the graph of its probability density function f consists of two straight line segments meeting at the point \(( 1 , k )\), as shown in the diagram. Find \(k\) and hence show that the distribution function F is given by $$\mathrm { F } ( x ) = \begin{cases} 0 & x \leqslant 0 , \\ \frac { 1 } { 3 } x ^ { 2 } & 0 < x \leqslant 1 , \\ x - \frac { 1 } { 2 } - \frac { 1 } { 6 } x ^ { 2 } & 1 < x \leqslant 3 , \\ 1 & x > 3 . \end{cases}$$ The random variable \(Y\) is given by \(Y = X ^ { 2 }\). Find
  1. the probability density function of \(Y\),
  2. the median value of \(Y\).

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The continuous random variable $X$ takes values in the interval $0 \leqslant x \leqslant 3$ only. For $0 \leqslant x \leqslant 3$ the graph of its probability density function f consists of two straight line segments meeting at the point $( 1 , k )$, as shown in the diagram. Find $k$ and hence show that the distribution function F is given by

$$\mathrm { F } ( x ) = \begin{cases} 0 & x \leqslant 0 , \\ \frac { 1 } { 3 } x ^ { 2 } & 0 < x \leqslant 1 , \\ x - \frac { 1 } { 2 } - \frac { 1 } { 6 } x ^ { 2 } & 1 < x \leqslant 3 , \\ 1 & x > 3 . \end{cases}$$

The random variable $Y$ is given by $Y = X ^ { 2 }$. Find\\
(i) the probability density function of $Y$,\\
(ii) the median value of $Y$.

\hfill \mbox{\textit{CAIE FP2 2011 Q11 OR}}