If the probability mass function (p.m.f.) of a random variable $X$ is $P(X=x) = \frac{1}{10}$ for $x = 1, 2, 3, \ldots, 10$,and $0$ otherwise,then $\operatorname{Var}(X)$ is equal to:

For the following probability distribution,find the expected value of $X$:

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$P(x)$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

$A$ random variable $X$ takes values $-1, 0, 1, 2$ with probabilities $\frac{1+3p}{4}, \frac{1-p}{4}, \frac{1+2p}{4}, \frac{1-4p}{4}$ respectively,where $p$ varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of $X$ are respectively.

If $X$ is a Poisson variate with $P(X=0)=0.8$, then the variance of $X$ is

For the following probability distribution, the standard deviation of the random variable $X$ is:
$X: 0, 1, 2$
$P(X): 0.3, 0.4, 0.3$

$A$ random variable $X$ has the following probability distribution:

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$3k$	$3k^2$	$k^2$	$2k^2$	$7k^2+k$

Determine $P(X > 6)$.