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The p.d.f. of c.r.v. $X$ is given by $f(x)=\frac{x+2}{18}$ if -2 < x < 4=0 otherwise then $\mathrm{P}[|x| < 1]=$
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The correct answer is:
$\frac{2}{9}$
(A)
$\begin{array}{c}
f(x)=\frac{x+2}{18}, \quad-2 < x < 4 \\
=0, \quad \text { otherwise } \\
\begin{aligned}
P(|x| < 1)=P(-1 < x < 1) &=\int_{-1}^{1} \frac{x+2}{18} \mathrm{dx} \\
&=\int_{-1}^{1} f(x) d x=\frac{1}{18}\left[\frac{1}{2}+2-\frac{1}{2}+2\right]=\frac{4}{18}=\frac{2}{9}
\end{aligned}
\end{array}$
$\begin{array}{c}
f(x)=\frac{x+2}{18}, \quad-2 < x < 4 \\
=0, \quad \text { otherwise } \\
\begin{aligned}
P(|x| < 1)=P(-1 < x < 1) &=\int_{-1}^{1} \frac{x+2}{18} \mathrm{dx} \\
&=\int_{-1}^{1} f(x) d x=\frac{1}{18}\left[\frac{1}{2}+2-\frac{1}{2}+2\right]=\frac{4}{18}=\frac{2}{9}
\end{aligned}
\end{array}$
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