The probability distribution of a random variable $X$ is as follows. Then the mean of $X$ is

$X = x_{i}$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x_{i})$	$k^2 / 3$	$k^2$	$2k^2 / 3$	$k / 2$	$k / 2$

For the following probability distribution,find the $Var(X)$.

$X$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X)$	$0.1$	$0.2$	$0.2$	$0.3$	$0.15$	$0.05$

(Given : $(0.25)^2 = 0.0625$,$(0.35)^2 = 0.1225$,$(0.45)^2 = 0.2025$)

The p.d.f. of a continuous random variable $X$ is given by $f(x) = \frac{1}{2}$ if $0 < x < 2$ and $f(x) = 0$ otherwise. If $a = P(X < \frac{1}{2})$ and $b = P(X > \frac{3}{2})$,then the relation between $a$ and $b$ is:

The cumulative distribution function (c.d.f.) $F(x)$ associated with the probability density function (p.d.f.) $f(x) = 3(1 - x^2)$ for $0 < x < 1$ and $f(x) = 0$ otherwise,is given by $F(x) = k(x - \frac{2x^3}{k})$. Find the value of $k$.

If $X$ is a Poisson variate such that $\frac{5}{3} k = P(X=2) = P(X=3)$,then $P(X=5) =$

$A$ die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die,then $P(E)$ is equal to: