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 piecewise PDF question requiring integration to find k, sketching, deriving the CDF, and finding the median. All steps are routine applications of core probability techniques with straightforward algebra. The piecewise nature adds mild complexity but no novel insight is required—this is slightly easier than average for Further Maths statistics.
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]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-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
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

## Question 4(a):

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

## Question 4(b):

| $\frac{kx^4}{4}$ 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):

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

## Question 4(d):

| $\int_0^1 kx^3\,dx + \int_1^m k(5-x)\,dx = \frac{1}{2}$ | 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]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-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]}}

This paper (5 questions)

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Q2 9 Q3 8 Q4 10 Q5 9 Q6 8

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

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

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