The p.d.f. of a continuous r.v. $\mathrm{X}$ is given by $\mathrm{f}(x)=\frac{x}{8}, 0

Question

The p.d.f. of a continuous r.v. $\mathrm{X}$ is given by $\mathrm{f}(x)=\frac{x}{8}, 0 < x < 4$ $=0$, otherwise, then $\mathrm{P}(\mathrm{X} \leq 2)$ is

Aicharya · Accepted Answer

$\begin{aligned} \mathrm{P}(\mathrm{x} \leq 2) &=\int_{0}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx} \ &=\int_{0}^{2} \frac{\mathrm{x}}{8} \mathrm{dx}=\frac{1}{8}\left[\frac{\mathrm{x}^{2}}{2}ight]_{0}^{2} \Rightarrow \frac{2^{2}}{16}-0=\frac{4}{16}=\frac{1}{4} \end{aligned}$

The p.d.f. of a continuous r.v. $\mathrm{X}$ is given by $\mathrm{f}(x)=\frac{x}{8}, 0 < x < 4$ $=0$,…

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