4.4 Random Variables

Q1

Topic	4.4 Random Variables
Tag
Source	M17/5/MATHL/HP1/ENG/TZ1/XX/10
Question Text	The continuous random variable $X$ has a probability density function given by $f(x)=\left\{k \sin \left(\frac{\pi x}{12}\right), 0 \leq x \leq 12\right\}$ (a) Find the value of $a$ , if $k=\frac{\pi}{a}$ where $a \in \mathbb{Z}^{+}$
Total Mark	4
Correct Answer	24
Explanation	na
Mark Scheme	$\begin{aligned} & k \int_0^{12} \sin \left(\frac{\pi x}{12}\right) d x=1 \\ & k\left[-\frac{12}{\pi} \cos \left(\frac{\pi x}{12}\right)\right]_0^{12}=1 \\ & k\left(\frac{12}{\pi}+\frac{12}{\pi}\right)=1 \end{aligned}$ Rearranging gives $k=\frac{\pi}{24}$ Answer: 24
Question Text	(b) By considering the graph of $f(x)$ write down (i) the mean of $X$ ; Total Mark: 1 Correct Answer: 6 Explanation: na Mark Scheme: na (ii) the median of $X$ ; Total Mark: 1 Correct Answer: 6 Explanation: na Mark Scheme: na (iii) the mode of $X$ Total Mark: 1 Correct Answer: 6 Explanation: na Mark Scheme: na
Question Text	(c) State the interquartile range of $X$ .
Total Mark	6
Correct Answer	4
Explanation	na
Mark Scheme	Tip 1: To find the first quartile, set the area as $\frac{1}{4}$ $\begin{gathered} \frac{\pi}{24} \int_0^n \sin \left(\frac{\pi x}{12}\right) d x=\frac{\pi}{24}\left[-\frac{12}{\pi} \cos \left(\frac{\pi x}{12}\right)\right]_0^n \\ \frac{\pi}{24}\left[-\frac{12}{\pi}\left(\cos \left(\frac{n \pi}{12}\right)-1\right)\right]=\frac{1}{4} \\ -\frac{1}{2} \cos \left(\frac{n \pi}{12}\right)+\frac{1}{2}=\frac{1}{4} \\ \cos \left(\frac{n \pi}{12}\right)=\frac{1}{2} \\ n=4 \end{gathered}$ So, $\begin{gathered} Q_1=4, Q_3=8 \\ I Q R=4 \end{gathered}$ Answer: 4
Question Text	(d) Calculate $P(X ≥ 4)$ (a) 0.7 (b) 0.75 (c) 0.8 (d) 0.85
Total Mark	1
Correct Answer	b
Explanation	na
Mark Scheme	$P(X ≥ 4)$ = 0.75 Answer: B

Q2

Topic	4.4 Random Variables
Tag	Random variables Discrete Expectation Variance Continuous Mode Median Mean Standard deviation Probability density function Probability mass function Linear Transformations Interquartile Range
Source	N14/5/MATHL/HP1/ENG/TZ0/XX/9
Question Text	A continuous random variable $T$ has probability density function $f$ defined by $f(t)=\{\|3-t\|, 2 \leq t \leq 4\}$ The interquartile range of $T$ can be written as $a-\sqrt{b}$ where $a, b \in Z^{+}$ . Find the value of $a+b$ .
Total Mark	5
Correct Answer	4
Explanation	na
Mark Scheme	$\int_2^{q_1}\|3-t\| d t=\frac{1}{4}$ As $q_1<3$ $\left[3 t-\frac{t^2}{2}\right]_2^{q_1}=\frac{1}{4}$ $\begin{gathered} 3 \times q_1-\frac{q_1^2}{2}-6+2=\frac{1}{4} \\ 2 q_1^2-12 q_1+17=0 \\ \frac{12 \pm \sqrt{144-136}}{4}=\frac{12 \pm \sqrt{8}}{4}=\frac{6 \pm \sqrt{2}}{2} \end{gathered}$ As $q_1>3$ $q_1=\frac{6+\sqrt{2}}{2}$ By symmetry $\begin{gathered} q_3=\frac{10-\sqrt{2}}{2} \\ q_3-q_1=\frac{4-2 \sqrt{2}}{2}=2-\sqrt{2} \end{gathered}$ Answer: 4

Q3

Topic	4.4 Random Variables
Tag
Source	M19-TZ1-P1-4(HL)
Question Text	The probability density function of the random variable X is defined as $f(x)=\left\{\cos x, 0 \leq x \leq \frac{\pi}{2}\right\}$ Find $\mathrm{E}(\mathrm{X})$ . [multiple choice] (a) $\frac{1}{2}$ (b) 1 (c) $\frac{\pi}{4}$ (d) $\frac{\pi}{2}-1$
Total Mark	5
Correct Answer	d
Explanation	na
Mark Scheme	Integration by parts $\begin{gathered} \int_0^{\frac{\pi}{2}} x \cos x d x=[x \sin x]_0^{\pi / 2}-\int_0^{\frac{\pi}{2}} \sin x d x \\ =[x \sin x+\cos x]_0^{\frac{\pi}{2}} \\ =\frac{\pi}{2}-1 \end{gathered}$ Answer: D

Q4

Topic	4.4 Random Variables
Tag	Random variables Continuous Probability density function
Source	M19/5/MATHL/HP1/ENG/TZ2/XX/10
Question Text	The random variable X has probability density function f given by $f(x)$ = {k(π - $arccos⁡x$ ) 0 ≤ x ≤1 where $k$ is a positive constant. (a) State the mode of X .
Total Mark	1
Correct Answer	1
Explanation	na
Mark Scheme	arccos1=0 Answer: 1
Question Text	(b) By considering the value of $∫$ arcos⁡xdx find the value of k (a) $\frac{1}{\pi -1}$ (b) $\frac{1}{\pi -2}$ (c) $\frac{2}{\pi -1}$ (d) $\frac{2}{\pi -2}$
Total Mark	1
Correct Answer	a
Explanation	na
Mark Scheme	attempt at integration by parts $\begin{gathered} \frac{d u}{d x}=-\frac{1}{\sqrt{1-x^2}}, \frac{d v}{d x}=1 \\ \int \operatorname{arcos} x d x=x \arccos x+\int \frac{x}{\sqrt{1-x^2}} d x \\ =x \arccos x-\sqrt{1-x^2}+c \\ \int_0^1(\pi-\arccos x) d x=\left[\pi x-x \arccos x+\sqrt{1-x^2}\right]_0^1 \\ =\pi-1 \end{gathered}$ As $\int_0^1 k(\pi-\arcsin x) d x=1$ $\begin{gathered} k(\pi-1)=1 \\ k=\frac{1}{\pi-1} \end{gathered}$ Answer: A