The p.m.f of a random variable $X$ is $P(X=x)=\frac{1}{2^5}\binom{5}{x}$,where $x=0, 1, 2, 3, 4, 5$ and $P(X=x)=0$ otherwise. Then:

The p.d.f. of a discrete random variable $X$ is defined as $f(x) = \begin{cases} kx^2, & x \in \{0, 1, 2, 3, 4, 5, 6\} \\ 0, & \text{otherwise} \end{cases}$. Then the value of $F(4)$ (c.d.f.) is:

If $f(x) = \frac{x}{8}$ for $0 < x < 4$ and $f(x) = 0$ otherwise,is the probability density function (p.d.f.) of a continuous random variable $X$,and $F(x)$ is the cumulative distribution function (c.d.f.) associated with $f(x)$,then find $F(0.5)$.

Two numbers are selected at random from the first six positive integers. If $X$ denotes the larger of the two numbers,then $\operatorname{Var}(X) = $

Let the mean and the standard deviation of the probability distribution be $\mu$ and $\sigma$,respectively. If $\sigma - \mu = 2$,then $\sigma + \mu$ is equal to:

$X$	$\alpha$	$1$	$0$	$-3$
$P(X)$	$\frac{1}{3}$	$K$	$\frac{1}{6}$	$\frac{1}{4}$

If the probability distribution of a random variable $X$ is as follows,then $P(X \leq 2) = $

$x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$3K$	$5K$	$3k^2$	$4k^2 + k$	$3k^2$