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…

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
  1. $\frac{5}{9}$
  2. $\frac{4}{9}$
  3. $\frac{2}{9}$
  4. $\frac{1}{9}$

Solution

$\begin{aligned} \mathrm{P}(|\mathrm{x}| < 2) &=\int_{-2}^{2} \frac{\mathrm{x}+2}{18} \mathrm{dx}=\frac{1}{18}\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-2}^{2} \\ &=\frac{1}{18}[(2-2)+2(2+2)]=\frac{8}{18}=\frac{4}{9} \end{aligned}$

Asked in: MHT CET 2020 (15 Oct Shift 1)

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