If a random variable $X$ has the following probability distribution, then the mean of $X$ is

$X = x_i$	$1$	$2$	$3$	$5$
$P(X = x_i)$	$2k^2$	$k$	$k$	$k^2$

If a random variable $X$ has the p.d.f. $f(x) = \begin{cases} \frac{k}{x^2+1} & , \text{if } 0 < x < \infty \\ 0 & , \text{otherwise} \end{cases}$,then the c.d.f. of $X$ is:

The probability distribution of a random variable $X$ is given below:

$X=x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$K$	$2K$	$2K$	$3K$	$K^2$	$2K^2$	$7K^2+K$

Then,$P(0 < X < 5)$ is equal to:

Suppose that a random variable $X$ follows a Poisson distribution. If $P(X=1) = P(X=2)$, then $P(X=5)$ is equal to:

The number of persons joining a cinema ticket counter in a minute follows a Poisson distribution with parameter $\lambda = 6$. The probability that at least one and at most five persons join the queue in a particular minute is:

If a discrete random variable $X$ is defined as follows:
$P(X=x) = \begin{cases} \frac{k(x+1)}{5^x}, & x=0, 1, 2, \ldots \\ 0, & \text{otherwise} \end{cases}$
then $k=$