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	Piecewise PDF with k
Difficulty	Standard +0.3 This is a standard Further Maths probability question requiring integration of a piecewise PDF to find k, then finding the CDF and median. While it involves multiple parts and careful handling of the piecewise structure, the techniques are routine: integrating polynomials, setting total probability to 1, and solving F(m)=0.5. The integration is straightforward and the question follows a predictable template for this topic.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03e Find cdf: by integration 5.03f Relate pdf-cdf: medians and percentiles

4 The continuous random variable $X$ has probability density function f given by $$f ( x ) = \begin{cases} k x ^ { 3 } & 0 \leqslant x < 1 , \\ k ( 5 - x ) & 1 \leqslant x \leqslant 5 , \\ 0 & \text { otherwise } , \end{cases}$$ where $k$ is a constant.

Sketch the graph of f.
Show that $k = \frac { 4 } { 33 }$. \includegraphics[max width=\textwidth, alt={}, center]{e2a45d19-7d48-4aa5-93f9-6ef90f99d7c4-09_2725_35_99_20}
Find the cumulative distribution function of $X$.
Find the median value of $X$.

Show mark scheme Show mark scheme source

Question 4(a):

Answer	Marks	Guidance
Answer	Marks	Guidance
Correct sketch	B1	Label 1 and 5 on $x$-axis, ignore labels, if any, on $y$-axis

Question 4(b):

Answer	Marks	Guidance
Answer	Marks	Guidance
$\frac{kx^4}{4}$ with limits 0 and 1 gives $\frac{k}{4}$; $k\left(5x - \frac{x^2}{2}\right)$ with limits 1 and 5 gives $8k$	B1	For either part correct
$\frac{k}{4} + 8k = 1$, $k = \frac{4}{33}$	B1	AG

Question 4(c):

Answer	Marks	Guidance
Answer	Marks	Guidance
$F(x) = \begin{cases} 0 & x < 0 \\ \frac{1}{33}x^4 & 0 \leq x < 1 \\ \frac{4}{33}\left(5x - \frac{1}{2}x^2 - \frac{17}{4}\right) & 1 \leq x \leq 5 \\ 1 & x > 5 \end{cases}$	M1	Integrate both parts
A1	Middle two parts correct AEF
A1	0 and 1 correct

Question 4(d):

Answer	Marks	Guidance
Answer	Marks	Guidance
$\int_0^1 kx^3\,dx + \int_1^m k(5-x)\,dx = \frac{1}{2}$	M1*	Or integrate $m$ to 5. Or use $F(x) = 0.5$
$\frac{4}{33}\left[\frac{1}{4} + 5m - \frac{1}{2}m^2 - 5 + \frac{1}{2}\right] = \frac{1}{2}$ or $\frac{4}{33}\left[\frac{25}{2} - 5m + \frac{1}{2}m^2\right] = \frac{1}{2}$ or $\frac{4}{33}\left[5m - \frac{1}{2}m^2 - \frac{17}{4}\right] = \frac{1}{2}$	M1	Must be a quadratic in $m$
$4m^2 - 40m + 67 = 0$	*DM1
$m = \frac{1}{2}(10 - \sqrt{33}) = 2.13$	A1

## Question 4(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct sketch | B1 | Label 1 and 5 on $x$-axis, ignore labels, if any, on $y$-axis |

## Question 4(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{kx^4}{4}$ with limits 0 and 1 gives $\frac{k}{4}$; $k\left(5x - \frac{x^2}{2}\right)$ with limits 1 and 5 gives $8k$ | B1 | For either part correct |
| $\frac{k}{4} + 8k = 1$, $k = \frac{4}{33}$ | B1 | AG |

## Question 4(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(x) = \begin{cases} 0 & x < 0 \\ \frac{1}{33}x^4 & 0 \leq x < 1 \\ \frac{4}{33}\left(5x - \frac{1}{2}x^2 - \frac{17}{4}\right) & 1 \leq x \leq 5 \\ 1 & x > 5 \end{cases}$ | M1 | Integrate both parts |
| | A1 | Middle two parts correct AEF |
| | A1 | 0 and 1 correct |

## Question 4(d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 kx^3\,dx + \int_1^m k(5-x)\,dx = \frac{1}{2}$ | M1* | Or integrate $m$ to 5. Or use $F(x) = 0.5$ |
| $\frac{4}{33}\left[\frac{1}{4} + 5m - \frac{1}{2}m^2 - 5 + \frac{1}{2}\right] = \frac{1}{2}$ or $\frac{4}{33}\left[\frac{25}{2} - 5m + \frac{1}{2}m^2\right] = \frac{1}{2}$ or $\frac{4}{33}\left[5m - \frac{1}{2}m^2 - \frac{17}{4}\right] = \frac{1}{2}$ | M1 | Must be a quadratic in $m$ |
| $4m^2 - 40m + 67 = 0$ | *DM1 | |
| $m = \frac{1}{2}(10 - \sqrt{33}) = 2.13$ | A1 | |

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Show LaTeX source

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

$$f ( x ) = \begin{cases} k x ^ { 3 } & 0 \leqslant x < 1 , \\ k ( 5 - x ) & 1 \leqslant x \leqslant 5 , \\ 0 & \text { otherwise } , \end{cases}$$

where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of f.
\item Show that $k = \frac { 4 } { 33 }$.\\

\includegraphics[max width=\textwidth, alt={}, center]{e2a45d19-7d48-4aa5-93f9-6ef90f99d7c4-09_2725_35_99_20}
\item Find the cumulative distribution function of $X$.
\item Find the median value of $X$.
\end{enumerate}

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

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Q1 6 Q2 9 Q3 8 Q4 10 Q5 9 Q6 8