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
  • 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
    $\frac{1}{9 x^3}+\frac{1}{9}$

Answer

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