Mathematics | MHT CET 2020 | Discussion Page

The p.d.f of c.r.v X is given by $f(x)=\frac{x+2}{18}$ , if -2 < x < 4=0, otherwise =0, then P[|x| <1]=

$\frac{2}{9}$

$\frac{4}{9}$

$\frac{1}{9}$

$\frac{1}{18}$

Option A

$f(x)=\begin{cases}\frac{x+2}{18}, & -2<x<4 = 0\\0 & otherwise\end{cases}$

$P[|x| <1]=\int_{-1}^{1 }f(x) dx$

$=\int_{-1}^{1 }\left(\frac{x+2}{18}\right)dx=\frac{1}{18}\left[\frac{x^{2}}{2}+2x\right]^{1}_{-1}$

$=\frac{1}{18}\left[\left(\frac{1}{2}+2\right)-\left(\frac{1}{2}-2\right)\right]=\frac{4}{18}=\frac{2}{9}$

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