Question

If the p.d.f of a continuous random variable $x$ is $f\left(x\right)=\{\begin{array}{cc}k{x}^2(1-x)& ,0<x<1\\ 0& ,\text{otherwise}\end{array}$

then value of $k$ is

• A. $12$
• B. $10$
• C. $-12$
• D. $8$

Answer 1

The correct option is A $12$

The pdf of a continuous random variable must satisy following condition:

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

$\Rightarrow {\int }_{-\infty }^{0}f\left(x\right)dx$ +${\int }_0^{1}f\left(x\right)dx+$${\int }_1^{\infty }f\left(x\right)dx=1$

$\Rightarrow$${\int }_0^{1}f\left(x\right)dx=1$

$\Rightarrow {[\frac{k{x}^3}{3}-\frac{k{x}^4}{4}]}_0^1=1$

$\Rightarrow k(\frac{1}{3}-\frac{1}{4})=1$

$\Rightarrow k=12$