CDF given as:
$$ F(x)= \begin{cases} \dfrac{1}{4}&\text{ if }& 0\leq x<\dfrac{1}{4}\\ \dfrac{1}{2}&\text{ if }&\dfrac{1}{4}\leq x<\dfrac{1}{2}\\ \dfrac{3}{4}&\text{ if }&\dfrac{1}{2}\leq x<\dfrac{3}{4}\\ \dfrac{x+3}{5}&\text{ if }&\dfrac{3}{4}\leq x<2\\ 1 &\text{ if }& x\geq2. \end{cases}$$
Then I tried to calculate PDF as given below: $$ f(x)= \begin{cases} \dfrac{1}{4}&\text{ if }& 0\leq x<\dfrac{1}{4}\\ \dfrac{1}{4}&\text{ if }&\dfrac{1}{4}\leq x<\dfrac{1}{2}\\ \dfrac{1}{4}&\text{ if }&\dfrac{1}{2}\leq x<\dfrac{3}{4}\\ \dfrac{1}{5}&\text{ if }&\dfrac{3}{4}\leq x<2\\ \end{cases}$$
I am having problem in last segment of PDF. Please someone tell me. Will it be zero? And how am I suppose to find cdf as $1$ out of it?
There is no pdf. The distribution is a mixture of a continuous part and that of a discrete part.
The discrete part is $\frac14,\frac14,\frac14$ at $0$, $\frac14$, $\frac12$, respectively. The continuous part is
$$g(x)=\begin{cases}\frac15&\text{ if }&\frac34<x<2\\0&\text{ otherwise.}\end{cases}$$
The following figure depicts this distribution: