If $f(x)=\frac{x}{8}$, if $0 < x < 4$ $=0$, otherwise is probability density function (p.d.f) of c.r.v. $X$…
If $f(x)=\frac{x}{8}$, if $0 < x < 4$
$=0$, otherwise is probability density function (p.d.f) of c.r.v. $X$ and $F(x)$ is c.d.f. associated with $f(x)$, then $\mathrm{F}(0 \cdot 5)=$
$\frac{1}{64}$
$\frac{1}{8}$
$\frac{1}{32}$
$\frac{1}{128}$
Solution
$F(x)=\int_{0}^{x} f(y) d y=\int_{0}^{x} \frac{y}{8} d y=\frac{1}{8}\left[\frac{y^{2}}{2}\right]_{0}^{x}$
$=\frac{1}{16}\left(x^{2}-0\right)=\frac{x^{2}}{16}$
$\therefore F(0.5)=\frac{(0.5)^{2}}{16}=\frac{1}{4 \times 16}=\frac{1}{64}$