Pearson Edexcel IAL Mathematics S2.2 Continuous random variables Question Bank

Practise working with continuous random variables through density functions, distribution functions, moments, sketches and quantile calculations.

Exam points

use areas under f(x) and values of F(x) to calculate probabilities over intervals
find constants, means and variances by integrating density functions over their support
interpret graphs of f(x) to locate modes, medians and quartiles from equations or inequalities

Covered syllabus points

S2.2.1 - concept of a continuous random variable S2.2.2 - Probability density and cumulative distribution functions S2.2.3 - Relationship between density and dF(x) f(x) =S2.2.4 - Mean and variance of continuous random variables S2.2.5 - Mode, median and quartiles of continuous random variables

[Maximum number: 15]

A random variable X has probability density function given by

f(x)=\left\{\begin{array}{cc} \frac{1}{4} & -\frac{1}{2} \leqslant x<\frac{1}{2} \\ 2 x-\frac{3}{4} & \frac{1}{2} \leqslant x \leqslant k \\ 0 & \text { otherwise } \end{array}\right.

where k is a positive constant.

(a)

Sketch the graph of f(x)

[ 2 ]

(b)

By forming and solving an equation in k, show that k=1.25

[ 4 ]

Area $=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+2 k-\frac{3}{4}\right)\left(k-\frac{1}{2}\right)=1$ or $\frac{1}{4}\left(k+\frac{1}{2}\right)+\frac{1}{2}(2 k-1)\left(k-\frac{1}{2}\right)=1$ or
$\frac{1}{4}+\left[\left(k^{2}-\frac{3}{4} k\right)-\left(\frac{1}{2}^{2}-\frac{3}{8}\right)\right]=1$

A1

$k=1.25^{*}$

A1 *

(c)

Use calculus to find E(X)

[ 4 ]

$\left[\int_{-0.5}^{0.5} \frac{1}{4} x \mathrm{~d} x\right]+\int_{0.5}^{1.25} 2 x^{2}-\frac{3}{4} x \mathrm{~d} x=[0]+\left[\frac{2 x^{3}}{3}-\frac{3}{8} x^{2}\right]_{0.5}^{1.25}$

M1A1

$=\left(\frac{2 \times 1.25^{3}}{3}-\frac{3}{8} \times 1.25^{2}\right)-\left(\frac{2 \times 0.5^{3}}{3}-\frac{3}{8} \times 0.5^{2}\right)$

dM1

$=\frac{93}{128}$

A1

(4)

(d)

Calculate the interquartile range of X

[ 5 ]

$Q_1=0.5$ $Q_3^2-\frac{3}{4}Q_3+0.125=0.5$ $Q_3=\frac{\frac34+\sqrt{\left(-\frac34\right)^2+4\left(\frac18\right)}}{2}=1.093\ldots$ $\operatorname{IQR}=1.093\ldots-0.5=0.593\ldots$ $\operatorname{IQR}=\text{awrt }0.59$
B1 for 0.5; M1 for equation for $Q_3$ ; M1 for solving; M1 for IQR; A1 for awrt 0.59.

[Maximum number: 8]

The continuous random variable X has cumulative distribution function given by

\mathrm{F}(x)=\left\{\begin{array}{lr} 0 & x<-k \\ \frac{x+k}{4 k} & -k \leqslant x \leqslant 3 k \\ 1 & x>3 k \end{array}\right.

where k is a positive constant.

(a)

Specify fully, in terms of k, the probability density function of X

[ 2 ]

f(x)=\begin{cases}\dfrac{1}{4k}, & -k\leqslant x\leqslant 3k,\\ 0, & \text{otherwise.}\end{cases}

(b)

Write down, in terms of k, the value of E(X)

[ 1 ]

\mathrm{E}(X)=k

(c)

Show that $\operatorname{Var}(X)=\frac{4}{3} k^{2}$

[ 2 ]

B1 M1 A1*

\operatorname{Var}(X)=\frac{(3k-(-k))^2}{12}=\frac{16k^2}{12}=\frac{4}{3}k^2

(d)

Find, in terms of k, the value of $\mathrm{E}(3X^2)$ .

[ 3 ]

M1 A1 A1

\mathrm{E}(X^2)=\operatorname{Var}(X)+\mathrm{E}(X)^2=\frac{4k^2}{3}+k^2=\frac{7k^2}{3}

\mathrm{E}(3X^2)=3\mathrm{E}(X^2)=7k^2

All question bank results loaded

Covered syllabus points

S2.2 - Continuous random variables question 1

Question (a)

Question (b)

Question (c)

Question (d)

S2.2 - Continuous random variables question 2

Question (a)

Question (b)

Question (c)

Question (d)