1. ## Discrete random variable

a discrete random variable X has the probability distribution given in the following table. it is given that E(X)=0.95

x ....... -1 ... 0 ... 1 ...2 ... 3
P(X=x) ...a...b...0.1...0.3...0.2...

Find the probability that X is greater than E(X)
Find values a and b
Calculate Var(X)

2. Originally Posted by gracey
a discrete random variable X has the probability distribution given in the following table. it is given that E(X)=0.95

x ....... -1 ... 0 ... 1 ...2 ... 3
P(X=x) ...a...b...0.1...0.3...0.2...

Find the probability that X is greater than E(X)
Find values a and b
Calculate Var(X)

Realize that finding the probability that X is greater than E(X) is equivalent to find $P(X>0.95)$, from given we know

$P(X>0.95)=P(1)+P(2)+P(3)=0.1+0.3+0.2=0.6$

Now let’s find a and b. By the definition of expected value, we have
$E(X)=\sum\limits_{x=-1}^3 xP(x)=(-1)P(-1)+(0)P(0)+(1)P(1)+(2)P(2)+(3)P(3)$
$=-a+0.1+0.6+0.6$

Since $E(X)=0.95$ we have $a=0.35$
Also by probability axiom, we have $\sum\limits_{x=-1}^3 P(x)=1$, In other words,

$a+b+0.1+0.3+0.2=1\Rightarrow b=0.05$

Lastly, the variance can be calculated by the formula $V(X)=E(X^2)-(E(X))^2$, note that
$E(X^2)=\sum\limits_{x=-1}^3 x^2P(x)$
$=(-1)^2P(-1)+(0)^2P(0)+(1)^2P(1)+(2)^2P(2)+(3)^2P(3)=3.45$

Hence $V(X)=E(X^2)-(E(X))^2=3.45-(0.95)^2=2.5475$

Roy