Question

The probability density function of a random variable X is given by :

f(x)={kx(2-x), 0<x<2

{0, e.w

a) Find the value of k

b)Find the distribution function F(X)

c)Find P(X>3)

Answer 1

(a) Probability density function of a random veriable $X$ is given by

$f(x)=kx(2-x)$ , $0<x<2$

$=0,$ e.w

We know that for $p.d.f$ ,

$\int _{-\infty}^{\infty} f(x)dx=1$

$\implies \int _ {0}^{2} kx(2-x)dx=1$

$\implies k.\int _ {0}^{2}(2x-x^2)dx=1$

$\implies k.[2.\frac{x^2}{2}-\frac {x^3}{3}]_{0}^{2}=1$

$\implies k.[4-\frac{8}{3}]=1$

$\implies k=\frac{3}{4}$

(b) The distribution function of the continuous random variable $X$ having probability density function $f(x)$ is given by ,

$F_X(x)=P(X\leq x)=\int _{-\infty}^{x}f(u) du$

Therefore for the given $p.d.f$ , the distribution function $F_X(x)=\intop_{-\infty}^{x}f(u)du$

$=\intop_{0}^{x}\frac{3}{4}u(2-u)du$

$=\frac{3}{4}\intop_{0}^{x}(2u-u^2)du$

$=\frac{3}{4}[u^2-\frac{u^3}{3}]_{0}^{x}$

$=\frac{1}{4}[3x^2-x^3]$

(c) $P(X>3)=\intop_{3}^{\infty}f(x)dx=0$ [ as $f(x)=0$ when $3<x<\infty$ ]