Question

If the p.d.f. of a continuous r.v. X is $f\left(x\right)\begin{array}{cc}=\frac{3}{20}x(x-2),& -1<x<4\\ =0,& otherwise\end{array}$ then the probability that X is negative, is?

• A. $1/5$
• B. $2/5$
• C. $3/5$
• D. $4/5$

Answer 1

The correct option is A $1/5$

Given, p.d.f is given by $f\left(x\right)=\{\frac{3}{20}x(x-2),-1<x<4\}$

$\{0,otherwise\}$

for c.d.f.

$\left(i\right)$ If $x\in (-\infty ,-1)$

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

$=0$

$\left(ii\right)$ If $x\in (-1,4)$

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

$={\int }_{-\infty }^{-1}0dx+{\int }_{-1}^x\frac{3}{20}(x^2-2x)dx$

$=\frac{3}{20}{[\frac{x^3}{3}-x^2]}_{-1}^{\sigma }$

$=\frac{3}{20}[\frac{x^3}{3}-x^2]-\left[\frac{3}{20}(\frac{-1}{3}-1)\right]$

$=\frac{3}{20}[\frac{x^3}{3}-x^2]+\frac{1}{5}$

$\left(iii\right)$ If $x\in (4,\infty ),f\left(x\right)=1$

$\Rightarrow p(x<0)=f\left(0\right)-f\left(0\right)$

$=\frac{1}{5}-0$

$=\frac{1}{5}$

Hence, the answer is $\frac{1}{5}.$