Question 7 - A-Level Maths

CAIE S2 2024 June — Question 7 10 marks

Exam Board	CAIE
Module	S2 (Statistics 2)
Year	2024
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	Single-piece PDF with k
Difficulty	Standard +0.3 This is a standard S2 probability density function question requiring routine integration techniques: (a) finding k using ∫f(x)dx=1, (b) numerical verification of the median by evaluating a definite integral, and (c) computing E(X) using integration by parts. All steps follow textbook procedures with no novel problem-solving required, making it slightly easier than average.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03c Calculate mean/variance: by integration 5.03f Relate pdf-cdf: medians and percentiles

7 The probability density function, f , of a random variable $X$ is given by $$f ( x ) = \begin{cases} k ( 1 + \cos x ) & 0 \leqslant x \leqslant \pi \\ 0 & \text { otherwise } \end{cases}$$ where $k$ is a constant.

Show that $k = \frac { 1 } { \pi }$.
Verify that the median of $X$ lies between 0.83 and 0.84 .
Find the exact value of $\mathrm { E } ( X )$.
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

Show mark scheme Show mark scheme source

Question 7(a):

Answer	Marks	Guidance
Answer	Mark	Guidance
$k\int_0^{\pi}(1 + \cos x)\,dx = 1$	M1	Attempt integrate $f(x)$ with correct limits and equate to 1.
$k\big[x + \sin x\big]_0^{\pi} = 1$	A1	Correct integration.
$k(\pi + \sin\pi - (0 + 0)) = 1$, $k\pi = 1$, $k = \frac{1}{\pi}$	A1	AG. Some evidence of substitution of limits, i.e. at least one interim step (e.g. $k\pi = 1$) as minimum requirement. Convincingly obtained; no errors seen.
3

Question 7(b):

Answer	Marks	Guidance
Answer	Mark	Guidance
$\frac{1}{\pi}\big[x + \sin x\big]_0^{0.83}$ or $\frac{1}{\pi}(0.83 + \sin 0.83)$ \	$\frac{1}{\pi}\big[x + \sin x\big]_0^{0.84}$ or $\frac{1}{\pi}(0.84 + \sin 0.84)$	M1
$= 0.499$ (3 sf) \	$= 0.504$ (3 sf)	A1
'$0.499$' $< 0.5 <$ '$0.504$' hence $0.83 <$ median $< 0.84$; Equivalent to $-0.000912 < 0 < 0.00441$ hence $0.83 <$ median $< 0.84$	A1FT	FT their areas; dep 0.5 is between their areas OE. If 0 scored, SC: $\frac{1}{\pi}(m + \sin m) = 0.5$ B1 and $m = 0.831$ to $0.832$, so $0.83 < m < 0.84$ B1.
3

Question 7(c):

Answer	Marks	Guidance
Answer	Mark	Guidance
$\frac{1}{\pi}\int_0^{\pi}(x + x\cos x)\,dx$	M1*	Attempt integrate $xf(x)$. Ignore limits.
$= \frac{1}{\pi}\left(\left[\frac{x^2}{2}\right]_0^{\pi} + \left[x\sin x\right]_0^{\pi} - \int_0^{\pi}(\sin x\,dx)\right)$	DM1	OE. Attempt to integrate (using 'parts') with correct limits, reaching an expression of the form $ax^2 + uv - \int v\,du$. OR using parts to integrate $x(1+\cos x)$ reaching an expression of the form $uv - \int v\,du$, i.e. $\frac{1}{\pi}(x^2 + x\sin x - \int(x + \sin x)\,dx)$
$= \frac{1}{\pi}\left(\frac{x^2}{2} + x\sin x + \cos x\right)$ or e.g. $\frac{\pi}{2} + \frac{1}{\pi}(0 - [-\cos x]_0^{\pi})$ or e.g. $\frac{\pi}{2} + \frac{1}{\pi}((-1-1))$	A1	Integration fully correct.
$= \frac{\pi}{2} - \frac{2}{\pi}$	A1	OE. ISW after correct exact value seen. SC1: Unsupported answer of $\frac{\pi}{2} - \frac{2}{\pi}$ scores B3. SC2: Unsupported answer of $0.934$ scores B2.
4

## Question 7(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $k\int_0^{\pi}(1 + \cos x)\,dx = 1$ | M1 | Attempt integrate $f(x)$ with correct limits and equate to 1. |
| $k\big[x + \sin x\big]_0^{\pi} = 1$ | A1 | Correct integration. |
| $k(\pi + \sin\pi - (0 + 0)) = 1$, $k\pi = 1$, $k = \frac{1}{\pi}$ | A1 | AG. Some evidence of substitution of limits, i.e. at least one interim step (e.g. $k\pi = 1$) as minimum requirement. Convincingly obtained; no errors seen. |
| | **3** | |

## Question 7(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{\pi}\big[x + \sin x\big]_0^{0.83}$ or $\frac{1}{\pi}(0.83 + \sin 0.83)$ \| $\frac{1}{\pi}\big[x + \sin x\big]_0^{0.84}$ or $\frac{1}{\pi}(0.84 + \sin 0.84)$ | M1 | Substitute correct limits into their integral. OR$_1$: integrate 0 to 0.83 and 0.84 to $\pi$. OR$_2$: use $g(m) = m + \sin m - (\pi/2)$ and find $g(0.83)$ and $g(0.84)$. OR$_3$: use $h(m) = m + \sin m$ and find $h(0.83)$ and $h(0.84)$. Both attempted. |
| $= 0.499$ (3 sf) \| $= 0.504$ (3 sf) | A1 | OR$_1$: 0.499 and 0.496. OR$_2$: $g(0.83) = -0.00286/7$ and $g(0.84) = 0.0138/9$. OR$_3$: $h(0.83) = 1.57$ and $h(0.84) = 1.58$ or $1.59$. Both correct. |
| '$0.499$' $< 0.5 <$ '$0.504$' hence $0.83 <$ median $< 0.84$; Equivalent to $-0.000912 < 0 < 0.00441$ hence $0.83 <$ median $< 0.84$ | A1FT | FT their areas; dep 0.5 is between their areas OE. If 0 scored, SC: $\frac{1}{\pi}(m + \sin m) = 0.5$ **B1** and $m = 0.831$ to $0.832$, so $0.83 < m < 0.84$ **B1**. |
| | **3** | |

## Question 7(c):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{\pi}\int_0^{\pi}(x + x\cos x)\,dx$ | M1* | Attempt integrate $xf(x)$. Ignore limits. |
| $= \frac{1}{\pi}\left(\left[\frac{x^2}{2}\right]_0^{\pi} + \left[x\sin x\right]_0^{\pi} - \int_0^{\pi}(\sin x\,dx)\right)$ | DM1 | OE. Attempt to integrate (using 'parts') with correct limits, reaching an expression of the form $ax^2 + uv - \int v\,du$. OR using parts to integrate $x(1+\cos x)$ reaching an expression of the form $uv - \int v\,du$, i.e. $\frac{1}{\pi}(x^2 + x\sin x - \int(x + \sin x)\,dx)$ |
| $= \frac{1}{\pi}\left(\frac{x^2}{2} + x\sin x + \cos x\right)$ or e.g. $\frac{\pi}{2} + \frac{1}{\pi}(0 - [-\cos x]_0^{\pi})$ or e.g. $\frac{\pi}{2} + \frac{1}{\pi}((-1-1))$ | A1 | Integration fully correct. |
| $= \frac{\pi}{2} - \frac{2}{\pi}$ | A1 | OE. ISW after correct exact value seen. SC1: Unsupported answer of $\frac{\pi}{2} - \frac{2}{\pi}$ scores **B3**. SC2: Unsupported answer of $0.934$ scores **B2**. |
| | **4** | |

Show LaTeX source

7 The probability density function, f , of a random variable $X$ is given by

$$f ( x ) = \begin{cases} k ( 1 + \cos x ) & 0 \leqslant x \leqslant \pi \\ 0 & \text { otherwise } \end{cases}$$

where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { \pi }$.
\item Verify that the median of $X$ lies between 0.83 and 0.84 .
\item Find the exact value of $\mathrm { E } ( X )$.\\

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

\hfill \mbox{\textit{CAIE S2 2024 Q7 [10]}}

This paper (7 questions)

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Q1 5 Q2 6 Q3 4 Q4 6 Q5 9 Q6 10 Q7 10

Answer	Marks	Guidance
Answer	Mark	Guidance
\(\frac{1}{\pi}\big[x + \sin x\big]_0^{0.83}\) or \(\frac{1}{\pi}(0.83 + \sin 0.83)\) \	\(\frac{1}{\pi}\big[x + \sin x\big]_0^{0.84}\) or \(\frac{1}{\pi}(0.84 + \sin 0.84)\)	M1
\(= 0.499\) (3 sf) \	\(= 0.504\) (3 sf)	A1
'\(0.499\)' \(< 0.5 <\) '\(0.504\)' hence \(0.83 <\) median \(< 0.84\); Equivalent to \(-0.000912 < 0 < 0.00441\) hence \(0.83 <\) median \(< 0.84\)	A1FT	FT their areas; dep 0.5 is between their areas OE. If 0 scored, SC: \(\frac{1}{\pi}(m + \sin m) = 0.5\) B1 and \(m = 0.831\) to \(0.832\), so \(0.83 < m < 0.84\) B1.
3