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

$X = x$	$1$	$2$	$3$	$\dots$	$n$
$P(X = x)$	$\frac{1}{n}$	$\frac{1}{n}$	$\frac{1}{n}$	$\dots$	$\frac{1}{n}$

Then $\operatorname{Var}(X) = $

The mean of the numbers obtained on throwing a die having written $1$ on three faces,$2$ on two faces,and $5$ on one face is

If the probability distribution of a discrete random variable $X$ is given by $P(X=k) = \frac{2^{-k}(3k+1)}{2^c}$ for $k = 0, 1, 2, \ldots, \infty$,then find the value of $P(X \leq c)$. Note: The expression provided in the prompt is likely $P(X=k) = \frac{(3k+1)}{2^{k+c}}$. Given $\sum_{k=0}^{\infty} P(X=k) = 1$,we have $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k+1) \left(\frac{1}{2}\right)^k = 1$.

In a Poisson distribution with unit mean,calculate the value of $\sum_{x=0}^{\infty} |x-\bar{x}| P(X=x)$,where $\bar{x}$ is the mean of the distribution.

In a Poisson distribution, if $P(X = 2)$ is twice $P(X = 1)$, then the standard deviation of the distribution is:

In a meeting,$60 \%$ of the members favour and $40 \%$ oppose a certain proposal. $A$ member is selected at random. We define a random variable $X$ such that $X=0$ if the member opposes the proposal and $X=1$ if the member is in favour. Then,$\text{Var}(X) = $