Question

A random variable X has probability density function f(x) as give below:
$f\left(x\right)=(\begin{array}{c}a+\mathrm{bx}\mathrm{for}0<x<1\\ 0\mathrm{otherwise}\end{array}$

If the expected value E[X] = 2/3, then Pr[X < 0.5] is .

• A. 0.25

Answer 1

The correct option is A 0.25
For p.d.f. ${\int }_{-\infty }^{\infty }f\left(x\right)dx=1$

$\Rightarrow {\int }_0^1(a+bx)dx=1$

$\Rightarrow {(ax+b\frac{x^2}{2})}_0^1=1$

$\Rightarrow a+\frac{b}{2}=1$

or $(2a+b)=2$ ...(1)

Now $E\left(X\right)=\frac{2}{3}\Rightarrow {\int }_{-\infty }^{\infty }xf\left(x\right)dx=\frac{2}{3}$

$\Rightarrow {\int }_0^{1}x(a+bx)dx=\frac{2}{3}$

$\Rightarrow (\frac{a{x}^2}{2}+\frac{b{x}^3}{2})=\frac{2}{3}$

$\Rightarrow (\frac{a}{2}+\frac{b}{2})=\frac{2}{3}$

$\Rightarrow (3a+2b)=4$ ...(2)

By solving (1) & (2) we get

$a=0$

and $b=2\mathrm{so}f\left(x\right)=\left(\begin{array}{cc}2x,& 0<x<1\\ 0,& \mathrm{otherwise}\end{array}\right\}$

So $P(X<0.5)={\int }_{-\infty }^{0.5}f\left(x\right)dx$

$={\int }_0^{0.5}\left(2x\right)dx=\frac{1}{4}=0.25$