The probability distribution of a random variable $X$ is given by the following table:

$X = x_i$	$3$	$5$	$7$	$9$
$P(X = x_i)$	$k$	$2k$	$3k$	$4k$

Then the standard deviation of $X$ is

If the range of a random variable $X$ is $\{0, 1, 2, 3, 4, \ldots\}$ with $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \geq 0$,then $a$ is equal to

$A$ random variable $X$ has the probability distribution
$\begin{array}{|c|c|c|c|c|c|c|}\hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\\hline P(X=x_i) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\\hline \end{array}$
If $A=\{x_i \mid x_i \text{ is a prime number}\}$ and $B=\{x_i \mid x_i < 4\}$ are two events,then $P(A \cup B) = $

Consider the probability distribution
$\begin{array}{|r|c|c|c|c|c|} \hline X=x & 1 & 2 & 3 & 4 & 5 \\ \hline P(X=x) & K & 2K & K^2 & 2K & 5K^2 \\ \hline \end{array}$
Then the value of $P(X > 2)$ is

If a discrete random variable $X$ has the probability distribution $P(X=x) = k \frac{2^{2x+1}}{(2x+1)!}$ for $x = 0, 1, 2, \ldots, \infty$,then $k =$

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

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

Then $P(X > 2)$ is equal to: