Let $p(x)$ represent the probability mass function of a Poisson distribution. If its mean $\lambda = 3.725$, then the value of $x$ at which $p(x)$ is maximum is

If the function $f$ defined by $f(x) = \begin{cases} K(x-x^2) & \text{if } 0 < x < 1 \\ 0 & \text{otherwise} \end{cases}$ is the probability density function (p.d.f.) of a random variable $X$,then the value of $P(X < \frac{1}{2})$ is

$A$ random variable $X$ has the range $\{0, 1, 2, \ldots\}$. If $P(X=r) = k(1+r) 3^{-r}$ for $r=0, 1, 2, \ldots$, where $k > 0$ is a real number, then $P(X=0) + P(X=1) + P(X=2) =$

If random variable $X$ is the waiting time in minutes for a bus and the probability density function of $X$ is given by $f(x) = \begin{cases} \frac{1}{5}, & 0 \leq x \leq 5 \\ 0, & \text{otherwise} \end{cases}$,then the probability of the waiting time being not more than $4$ minutes 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

For the probability distribution given below,find $\operatorname{Var}(X)$.

$X$	$5$	$6$	$7$	$8$	$9$	$10$	$11$
$P(X=x)$	$0.07$	$0.2$	$0.3$	$k$	$0.07$	$0.04$	$0.02$