Probability

For the distribution function given by $\;$ $F \left(x\right) = \begin{cases} 0, & x < 0 \\ & \\ x^2, & 0 \leq x \leq 1 \\ & \\ 1, & x > 1 \end{cases}$
find the density function. Also evaluate

1. $P \left(0.5 < x < 0.75\right)$


2. $P \left(x \leq 0.5\right)$


3. $P \left(x > 0.75\right)$

Probability density function (p.d.f) $= f \left(x\right) = \dfrac{d}{dx} F \left(x\right)$

Now, $\dfrac{d}{dx} \left(x^2\right) = 2 x$ $\;$ and $\;$ $\dfrac{d}{dx} \left(1\right) = 0$

$\therefore \;$ $f \left(x\right) = \begin{cases} 2 x, & 0 \leq x \leq 1 \\ & \\ 0, & \text{ elsewhere} \end{cases}$

1. $P \left(0.5 < x < 0.75\right) = \int \limits_{0.5}^{0.75} 2 \; x \; dx$
   
   Now,
   
   $\begin{aligned} \int \limits_{0.5}^{0.75} 2 \; x \; dx & = 2 \times \left[\dfrac{x^2}{2}\right]_{0.5}^{0.75} \\\\ & = \left(0.75\right)^2 - \left(0.5\right)^2 = 0.3125 \end{aligned}$
   
   $\therefore \;$ $P \left(0.5 < x < 0.75\right) = 0.3125$
   
   
2. $P \left(x \leq 0.5\right) = \int \limits_{0}^{0.5} 2 \; x \; dx$
   
   Now,
   
   $\begin{aligned} \int \limits_{0}^{0.5} 2 \; x \; dx & = 2 \times \left[\dfrac{x^2}{2}\right]_{0}^{0.5} \\\\ & = \left(0.5\right)^2 = 0.25 \end{aligned}$
   
   $\therefore \;$ $P \left(x \leq 0.5\right) = 0.25$
   
   
3. $P \left(x > 0.75\right) = \int \limits_{0.75}^{1} 2 \; x \; dx$
   
   Now,
   
   $\begin{aligned} \int \limits_{0.75}^{1} 2 \; x \; dx & = 2 \times \left[\dfrac{x^2}{2}\right]_{0.75}^{1} \\\\ & = \left(1\right)^2 - \left(0.75\right)^2 = 0.4375 \end{aligned}$
   
   $\therefore \;$ $P \left(x > 0.75\right) = 0.4375$