# Problem regarding cumulative distribution function

Let $X$ be a random variable with the following cumulative distribution function:

$$F(x)= \begin{cases} 0 & \quad x<0\\ x^2 & \quad 0\leq x<\frac{1}{2}\\ \frac{3}{4} & \quad \frac{1}{2}\leq x<1\\ 1 &\quad x\geq1. \end{cases}$$

Then what is the value of $P(\frac{1}{4}<X<1).$

I am trying to solve this problem as:

$P(\frac{1}{4}<X<1)=P(\frac{1}{4}<X\leq1)$, since for continuous random variable probability at a point is always zero. Thus $P(\frac{1}{4}<X<1)=P(\frac{1}{4}<X\leq1)=F(1)-F(\frac{1}{4})=\frac{15}{16}=0.9375$. But the answer didn't match. Answer is 0.68. So where did i wrong. Any suggestion or solution regarding this should be highly appreciated.

• Who told you the CDF is continuous at $x=1$? More care, please... – Did Jan 27 '18 at 8:57
• @Did I guess it as the range of the random variable $X$ is in piece-wise continuous form. – SAHEB PAL Jan 27 '18 at 17:12
• Stop guessing then, and start using solid definitions. Just my two cents. – Did Jan 27 '18 at 18:25

$$P\left(\dfrac{1}{4}<X<1\right)=F\left(1^-\right)-F\left(\dfrac{1}{4}^{+}\right)=\dfrac{3}{4}-\dfrac{1}{16}=\dfrac{11}{16}$$
• Using \left( and \right) delimiters might make formatting better. – TheSimpliFire Jan 27 '18 at 9:28
• $F(x)=F(x^+)$ for every $x$ and every CDF $F$. – Did Jan 30 '18 at 7:34
• That's why the inequalities in the question have been defined so. Also $F(x)$ $F(x^{-})$ and $F(x^+)$ are three distinct number generally and they are equal only if $F(x)$ is continuous. – Mostafa Ayaz Jan 30 '18 at 17:17