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Let $X$ be a continuous random variable with p.d.f:
$f(x)= \begin{cases}a x, & 0<x \leq 1 \\ a, & 1 \leq x \leq 2 \\ -a x+3 a, & 2 \leq x \leq 3 \\ 0, & \text { otherwise }\end{cases}$
(i) Find 'a'
(ii) compute $P(X \leq 1.5)$
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Since $f(x)$ is p.d.f $\int_{-\infty}^{\infty} f(x) d x=1$
\begin{aligned} &\Rightarrow \int_{0}^{1}(a x) d x+\int_{1}^{2} a d x+\int_{2}^{3}(-a x+3 a) d x=1 \\ &\left(\frac{a x^{2}}{2}\right)_{0}^{1}+(a x)_{1}^{2}+\left[\frac{-a x^{2}}{2}+3 a x\right]_{2}^{3}=1 \\ &\frac{a}{2}+2 a-a-\frac{9}{2} a+9 a+2 a-6 a=1 \\ &\frac{a}{2}-\frac{9 a}{2}+6 a=1 \\ &2 a=1 \\ &\Rightarrow a=\frac{1}{2} \\ &P(X \leq 1.5)=\int_{0}^{1} a x d x+\int_{1}^{1.5} a d x \\ &=\left(\frac{a x^{2}}{2}\right)_{0}^{1}+(a x)_{1}^{1.5} \\ &=\frac{a}{2}+1.5 a-\dot{a}=\frac{a}{2}+\frac{a}{2}=a \\ &=\frac{1}{2}\left(\text { since } a=\frac{1}{2}\right) \end{aligned}
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