If a continuous random variable $\mathrm{X}$ has probability density function $\mathrm{f}(x)$ given by…
If a continuous random variable $\mathrm{X}$ has probability density function $\mathrm{f}(x)$ given by
$f(x)=\left\{\begin{array}{cc}
a x & , \text { if } 0 \leq x < 1 \\
a & , \text { if } 1 \leq x < 2 \\
3 a-a x & \text { if } 2 \leq x \leq 3 \\
0 & \text {, otherwise }
\end{array},\right.$
then a has the value
$\frac{1}{5}$
$\frac{1}{3}$
$\frac{1}{2}$
1
Solution
Since $\mathrm{f}(x)$ is the p.d.f. of $\mathrm{X}$,
$\begin{aligned}
& \int_{-\infty}^{\infty} \mathrm{f}(x) \mathrm{d} x=1 \\
& \Rightarrow \int_0^1 \mathrm{a} x \mathrm{~d} x+\int_1^2 \mathrm{ad} x+\int_2^3(3 \mathrm{a}-\mathrm{ax}) \mathrm{d} x=1 \\
& \Rightarrow \mathrm{a}\left[\frac{x^2}{2}\right]_0^1+\mathrm{a}[x]_1^2+\left[3 \mathrm{a} x-\frac{\mathrm{a} x^2}{2}\right]_2^3=1
\end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{a}\left(\frac{1}{2}\right)+\mathrm{a}(1)+\left(\frac{9 \mathrm{a}}{2}-4 \mathrm{a}\right)=1 \\ & \Rightarrow 2 \mathrm{a}=1 \\ & \Rightarrow \mathrm{a}=\frac{1}{2}\end{aligned}$