Question 6 - A-Level Maths

CAIE Further Paper 4 2023 June — Question 6 11 marks

Exam Board	CAIE
Module	Further Paper 4 (Further Paper 4)
Year	2023
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	Find or specify CDF
Difficulty	Challenging +1.2 This is a multi-part Further Maths statistics question requiring integration of exponential functions to find the CDF, transformation of random variables using the Jacobian method, and calculation of percentiles and expectations. While it involves several techniques, each step follows standard procedures taught in Further Statistics courses without requiring novel insight or particularly complex manipulation.
Spec	5.03a Continuous random variables: pdf and cdf 5.03c Calculate mean/variance: by integration 5.03e Find cdf: by integration 5.03f Relate pdf-cdf: medians and percentiles 5.07a Non-parametric tests: when to use

6 The continuous random variable $X$ has probability density function f given by $$f ( x ) = \begin{cases} \frac { 3 } { 28 } \left( e ^ { \frac { 1 } { 2 } x } + 4 e ^ { - \frac { 1 } { 2 } x } \right) & 0 \leqslant x \leqslant 2 \ln 3 \\ 0 & \text { otherwise } \end{cases}$$

Find the cumulative distribution function of $X$.
The random variable $Y$ is defined by $Y = e ^ { \frac { 1 } { 2 } ( X ) }$.
Find the probability density function of $Y$.
Find the 30th percentile of $Y$.
Find $\mathrm { E } \left( Y ^ { 4 } \right)$.
If you use the following page to complete the answer to any question, the question number must be clearly shown.

Show mark scheme Show mark scheme source

Question 6(a):

Answer	Marks	Guidance
$F(x) = \frac{3}{28}\left(2e^{\frac{1}{2}x} - 8e^{-\frac{1}{2}x}(+c)\right)$	M1	$+c$ not required
$F(x) = \frac{3}{28}\left(2e^{\frac{1}{2}x} - 8e^{-\frac{1}{2}x}\right) + \frac{9}{14}$	A1
Correct ranges including $0 \leq x \leq 2\ln 3$ associated with their $F(x)$; $F(x) = 0$ for $x < 0$ and $1$ for $x > 2\ln 3$	B1	No gaps in range

Question 6(b):

Answer	Marks	Guidance
$G(y) = \frac{3}{28}\left(2y - \frac{8}{y} + 6\right)$	M1	$y$ substituted into their F
$1 \leq y \leq 3$	B1	Seen anywhere, condone $1 \leq y \leq e^{\ln 3}$
$g(y) = \frac{3}{28}\left(2 + \frac{8}{y^2}\right)$	A1	Correct expression, not containing logs

Question 6(c):

Answer	Marks	Guidance
$\frac{3}{28}\left(2y - \frac{8}{y} + 6\right) = \frac{3}{10}$	M1	Their $G(y) = \frac{3}{10}$
$5y^2 + 8y - 20 = 0$	M1	Obtain from $G(y)$ and solve quadratic equation to find $y$
$y = 1.35$	A1	Single correct answer

Question 6(d):

Answer	Marks	Guidance
\[E(Y^4) = \int_1^3 \frac{3}{28}\left(2+\frac{8}{y^2}\right)y^4\,dy = \int_1^3 \frac{3}{28}\left(2y^4+8y^2\right)dy = \frac{3}{28}\left[\frac{2}{5}y^5+\frac{8}{3}y^3\right]\]	M1	Integrate $y^4 \times their\ g(y)$. Limits must be 1 and 3.
$17.8$	A1	$\dfrac{89}{5}$

Total: 2 marks

## Question 6(a):

| $F(x) = \frac{3}{28}\left(2e^{\frac{1}{2}x} - 8e^{-\frac{1}{2}x}(+c)\right)$ | M1 | $+c$ not required |
|---|---|---|
| $F(x) = \frac{3}{28}\left(2e^{\frac{1}{2}x} - 8e^{-\frac{1}{2}x}\right) + \frac{9}{14}$ | A1 | |
| Correct ranges including $0 \leq x \leq 2\ln 3$ associated with their $F(x)$; $F(x) = 0$ for $x < 0$ and $1$ for $x > 2\ln 3$ | B1 | No gaps in range |

---

## Question 6(b):

| $G(y) = \frac{3}{28}\left(2y - \frac{8}{y} + 6\right)$ | M1 | $y$ substituted into their F |
|---|---|---|
| $1 \leq y \leq 3$ | B1 | Seen anywhere, condone $1 \leq y \leq e^{\ln 3}$ |
| $g(y) = \frac{3}{28}\left(2 + \frac{8}{y^2}\right)$ | A1 | Correct expression, not containing logs |

---

## Question 6(c):

| $\frac{3}{28}\left(2y - \frac{8}{y} + 6\right) = \frac{3}{10}$ | M1 | Their $G(y) = \frac{3}{10}$ |
|---|---|---|
| $5y^2 + 8y - 20 = 0$ | M1 | Obtain from $G(y)$ and solve quadratic equation to find $y$ |
| $y = 1.35$ | A1 | Single correct answer |

## Question 6(d):

$$E(Y^4) = \int_1^3 \frac{3}{28}\left(2+\frac{8}{y^2}\right)y^4\,dy = \int_1^3 \frac{3}{28}\left(2y^4+8y^2\right)dy = \frac{3}{28}\left[\frac{2}{5}y^5+\frac{8}{3}y^3\right]$$ | M1 | Integrate $y^4 \times their\ g(y)$. Limits must be 1 and 3.

$17.8$ | A1 | $\dfrac{89}{5}$

**Total: 2 marks**

Show LaTeX source

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

$$f ( x ) = \begin{cases} \frac { 3 } { 28 } \left( e ^ { \frac { 1 } { 2 } x } + 4 e ^ { - \frac { 1 } { 2 } x } \right) & 0 \leqslant x \leqslant 2 \ln 3 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find the cumulative distribution function of $X$.\\

The random variable $Y$ is defined by $Y = e ^ { \frac { 1 } { 2 } ( X ) }$.
\item Find the probability density function of $Y$.
\item Find the 30th percentile of $Y$.
\item Find $\mathrm { E } \left( Y ^ { 4 } \right)$.\\

If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 4 2023 Q6 [11]}}

This paper (6 questions)

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Q1 4 Q2 8 Q3 9 Q4 9 Q5 9 Q6 11

Answer	Marks	Guidance
\(F(x) = \frac{3}{28}\left(2e^{\frac{1}{2}x} - 8e^{-\frac{1}{2}x}(+c)\right)\)	M1	\(+c\) not required
\(F(x) = \frac{3}{28}\left(2e^{\frac{1}{2}x} - 8e^{-\frac{1}{2}x}\right) + \frac{9}{14}\)	A1
Correct ranges including \(0 \leq x \leq 2\ln 3\) associated with their \(F(x)\); \(F(x) = 0\) for \(x < 0\) and \(1\) for \(x > 2\ln 3\)	B1	No gaps in range

Answer	Marks	Guidance
\(G(y) = \frac{3}{28}\left(2y - \frac{8}{y} + 6\right)\)	M1	\(y\) substituted into their F
\(1 \leq y \leq 3\)	B1	Seen anywhere, condone \(1 \leq y \leq e^{\ln 3}\)
\(g(y) = \frac{3}{28}\left(2 + \frac{8}{y^2}\right)\)	A1	Correct expression, not containing logs

Answer	Marks	Guidance
\(\frac{3}{28}\left(2y - \frac{8}{y} + 6\right) = \frac{3}{10}\)	M1	Their \(G(y) = \frac{3}{10}\)
\(5y^2 + 8y - 20 = 0\)	M1	Obtain from \(G(y)\) and solve quadratic equation to find \(y\)
\(y = 1.35\)	A1	Single correct answer

Answer	Marks	Guidance
\[E(Y^4) = \int_1^3 \frac{3}{28}\left(2+\frac{8}{y^2}\right)y^4\,dy = \int_1^3 \frac{3}{28}\left(2y^4+8y^2\right)dy = \frac{3}{28}\left[\frac{2}{5}y^5+\frac{8}{3}y^3\right]\]	M1	Integrate \(y^4 \times their\ g(y)\). Limits must be 1 and 3.
\(17.8\)	A1	\(\dfrac{89}{5}\)