Question 4 - A-Level Maths

CAIE Further Paper 4 2024 November — Question 4 10 marks

Exam Board	CAIE
Module	Further Paper 4 (Further Paper 4)
Year	2024
Session	November
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	Find or specify CDF
Difficulty	Standard +0.8 This is a multi-part Further Maths statistics question requiring: (a) integration to find CDF, (b) transformation of random variables using the Jacobian method, (c) solving F_Y(m)=0.5 involving fourth roots, and (d) computing E(Y) via integration. The transformation in part (b) requires careful handling of the Jacobian and domain mapping, which goes beyond standard A-level. While each technique is accessible, the combination and the non-linear transformation elevate this above typical Further Stats questions.
Spec	5.03a Continuous random variables: pdf and cdf 5.03e Find cdf: by integration 5.03g Cdf of transformed variables

4 The random variable $X$ has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 21 } ( x - 1 ) ^ { 2 } & 2 \leqslant x \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$

Find the cumulative distribution function of $X$.
The random variable $Y$ is defined by $Y = ( X - 1 ) ^ { 4 }$.
Find the probability density function of $Y$. \includegraphics[max width=\textwidth, alt={}, center]{b9cbf607-4f40-41bb-8374-6b2c39f945ac-09_2725_35_99_20}
Find the median value of $Y$.
Find $\mathrm { E } ( Y )$.

Show mark scheme Show mark scheme source

Question 4(a):

Answer	Marks	Guidance
$F(X)=\begin{cases}0 & x<2,\\ \frac{1}{63}\left[(x-1^3-1)\right] & 2\leqslant x\leqslant 5,\\ 1 & x>5.\end{cases}$	M1	Integrate $\frac{1}{63}(x^3-3x^2+3x)[+c]$ oe.
A1	Correct $F(x)$ (with $c=-\frac{2}{63}$ evaluated).
A1	0 and 1 correct, inequalities correct.
3

Question 4(b):

Answer	Marks	Guidance
$G(y)=\frac{1}{63}\left(y^{0.75}-1\right)$	M1	Use transformation correctly on function. May see $\frac{1}{63}(y^{0.25}+1)^3-\frac{1}{21}(y^{0.25}+1)^2+\frac{1}{21}(y^{0.25}+1)-\frac{2}{63}$
$g(y)=\begin{cases}\frac{1}{84}y^{-0.25} & 1\leqslant y\leqslant 256,\\ 0 & \text{otherwise.}\end{cases}$	M1	Differentiate and change variable in domain.
A1	All correct and simplified.
3

Question 4(c):

Answer	Marks	Guidance
$\frac{1}{63}(y^{0.75}-1)=0.5$	M1	Equate their $G(y)$ to 0.5 and attempt to solve.
$y=104$	A1	AWRT 104.
2

Question 4(d):

Answer	Marks	Guidance
$\int_1^{256}\frac{1}{84}y(y^{-0.25})dy=\int_1^{256}\frac{1}{84}(y^{0.75})dy=\left[\frac{1}{84}\times\frac{4}{7}y^{1.75}\right]$	M1	Use their $y\,g(y)$, integrate, ignore limits.
$111$	A1	$\frac{5461}{49}$
2

## Question 4(a):

$F(X)=\begin{cases}0 & x<2,\\ \frac{1}{63}\left[(x-1^3-1)\right] & 2\leqslant x\leqslant 5,\\ 1 & x>5.\end{cases}$ | M1 | Integrate $\frac{1}{63}(x^3-3x^2+3x)[+c]$ oe.

| A1 | Correct $F(x)$ (with $c=-\frac{2}{63}$ evaluated).

| A1 | 0 and 1 correct, inequalities correct.

| 3 |

## Question 4(b):

$G(y)=\frac{1}{63}\left(y^{0.75}-1\right)$ | M1 | Use transformation correctly on function. May see $\frac{1}{63}(y^{0.25}+1)^3-\frac{1}{21}(y^{0.25}+1)^2+\frac{1}{21}(y^{0.25}+1)-\frac{2}{63}$

$g(y)=\begin{cases}\frac{1}{84}y^{-0.25} & 1\leqslant y\leqslant 256,\\ 0 & \text{otherwise.}\end{cases}$ | M1 | Differentiate and change variable in domain.

| A1 | All correct and simplified.

| 3 |

## Question 4(c):

$\frac{1}{63}(y^{0.75}-1)=0.5$ | M1 | Equate *their* $G(y)$ to 0.5 and attempt to solve.

$y=104$ | A1 | AWRT 104.

| 2 |

## Question 4(d):

$\int_1^{256}\frac{1}{84}y(y^{-0.25})dy=\int_1^{256}\frac{1}{84}(y^{0.75})dy=\left[\frac{1}{84}\times\frac{4}{7}y^{1.75}\right]$ | M1 | Use *their* $y\,g(y)$, integrate, ignore limits.

$111$ | A1 | $\frac{5461}{49}$

| 2 |

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4 The random variable $X$ has probability density function f given by

$$f ( x ) = \begin{cases} \frac { 1 } { 21 } ( x - 1 ) ^ { 2 } & 2 \leqslant x \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find the cumulative distribution function of $X$.\\

The random variable $Y$ is defined by $Y = ( X - 1 ) ^ { 4 }$.
\item Find the probability density function of $Y$.\\

\includegraphics[max width=\textwidth, alt={}, center]{b9cbf607-4f40-41bb-8374-6b2c39f945ac-09_2725_35_99_20}
\item Find the median value of $Y$.
\item Find $\mathrm { E } ( Y )$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q4 [10]}}

This paper (6 questions)

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Q1 4 Q2 8 Q3 10 Q4 10 Q5 9 Q6 9

Answer	Marks	Guidance
\(F(X)=\begin{cases}0 & x<2,\\ \frac{1}{63}\left[(x-1^3-1)\right] & 2\leqslant x\leqslant 5,\\ 1 & x>5.\end{cases}\)	M1	Integrate \(\frac{1}{63}(x^3-3x^2+3x)[+c]\) oe.
A1	Correct \(F(x)\) (with \(c=-\frac{2}{63}\) evaluated).
A1	0 and 1 correct, inequalities correct.
3

Answer	Marks	Guidance
\(G(y)=\frac{1}{63}\left(y^{0.75}-1\right)\)	M1	Use transformation correctly on function. May see \(\frac{1}{63}(y^{0.25}+1)^3-\frac{1}{21}(y^{0.25}+1)^2+\frac{1}{21}(y^{0.25}+1)-\frac{2}{63}\)
\(g(y)=\begin{cases}\frac{1}{84}y^{-0.25} & 1\leqslant y\leqslant 256,\\ 0 & \text{otherwise.}\end{cases}\)	M1	Differentiate and change variable in domain.
A1	All correct and simplified.
3

Answer	Marks	Guidance
\(\frac{1}{63}(y^{0.75}-1)=0.5\)	M1	Equate their \(G(y)\) to 0.5 and attempt to solve.
\(y=104\)	A1	AWRT 104.
2

Answer	Marks	Guidance
\(\int_1^{256}\frac{1}{84}y(y^{-0.25})dy=\int_1^{256}\frac{1}{84}(y^{0.75})dy=\left[\frac{1}{84}\times\frac{4}{7}y^{1.75}\right]\)	M1	Use their \(y\,g(y)\), integrate, ignore limits.
\(111\)	A1	\(\frac{5461}{49}\)
2