MCQ
If the p.d.f. of a continuous random variable $X$ is given as
$
f(x)=\frac{x^2}{3},-1$
$=0$, otherwise then c.d.f. of $X$ is
$
f(x)=\frac{x^2}{3},-1
$=0$, otherwise then c.d.f. of $X$ is
- A$\frac{x^3}{9}+\frac{1}{9}$
- B$\frac{x^3}{9}-\frac{1}{9}$
- C$\frac{x^3}{4}+\frac{1}{4}$
- D$\frac{1}{9 x^3}+\frac{1}{9}$