If $X$ is a random variable with the following probability distribution:

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

Then the variance of $X$ is:

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

$X = x$	$-1$	$0$	$1$	$2$
$P(X = x)$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$

Then the value of $6 \Sigma(x^2) P(X=x) - \operatorname{var}(X) =$ ?

If $3$ is the variance of a Poisson distribution,then $P(1 < x < 4) = $

The random variable $X$ has the following probability distribution:
| $X$ | $8$ | $12$ | $16$ | $20$ | $24$ |
|---|---|---|---|---|---|
| $P(X)$ | $K$ | $\frac{1}{6}$ | $\frac{3}{8}$ | $2K$ | $\frac{1}{12}$ |
Then the value of $K$ is:

For the probability distribution of a discrete random variable $X$ as given below,the mean of $X$ is:

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$K + \frac{2}{10}$	$K + \frac{3}{10}$	$K + \frac{3}{10}$	$K + \frac{4}{10}$	$K + \frac{2}{10}$

$A$ player tosses two coins. He wins $Rs. 10$ if $2$ heads appear,$Rs. 5$ if one head appears,and $Rs. 2$ if no head appears. Find the variance of the winning amount.