In a pizza hut,the following distribution is found for the daily demand of pizzas. Then the expected daily demand and variance are respectively:

No. of Pizzas $(x_i)$	$5$	$6$	$7$	$8$	$9$	$10$
Probability $(P_i)$	$0.07$	$0.2$	$0.3$	$0.3$	$0.07$	$0.06$

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

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(x)$	$0.01$	$0.10$	$0.26$	$0.33$	$0.18$	$0.06$	$K$	$0.04$

Then $P(X \geq 3) - P(X < 6) =$

The following is the probability distribution of $X$:

$X$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1+p}{5}$	$\frac{2-2p}{5}$	$\frac{2-p}{5}$	$\frac{2p}{5}$

For a minimum value of $p$,the value of $5 E(X)$ is:

If $X$ is a random variable with cumulative distribution function $F(x)$ and its probability distribution is given by the following table:

$X = x$	$-1.5$	$-0.5$	$0.5$	$1.5$	$2.5$
$P(X = x)$	$0.05$	$0.2$	$0.15$	$0.25$	$0.35$

Then,find the value of $F(1.5) - F(-0.5)$.

From a lot of $12$ items containing $3$ defectives,a sample of $5$ items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $n-m$ is equal to..........

State whether the following table represents a probability distribution of a random variable $X$. Give reasons for your answer.

$X$	$0$	$1$	$2$
$P(X)$	$0.4$	$0.4$	$0.2$