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If $f(x)=\frac{x+2}{18},-2 < x < 4$
$=0 \quad, \quad$ otherwise,
is the p. d. f. of a r. v. X, then the value of $\mathrm{P}(|\mathrm{X}| < 2)$ is
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$=0 \quad, \quad$ otherwise,
is the p. d. f. of a r. v. X, then the value of $\mathrm{P}(|\mathrm{X}| < 2)$ is
Solution:
1002 Upvotes
Verified Answer
The correct answer is:
$\frac{4}{9}$
$\begin{aligned} \mathrm{P}(|\mathrm{x}| < 2) &=\int_{-2}^{2} \frac{\mathrm{x}+2}{18} \mathrm{dx}=\frac{1}{18}\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-2}^{2} \\ &=\frac{1}{18}[(2-2)+2(2+2)]=\frac{8}{18}=\frac{4}{9} \end{aligned}$
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