# Discrete Random Variables

• April 13th 2010, 01:56 AM
BabyMilo
Discrete Random Variables
Given E[X]=0.95. Find the probability that X is greater than E[X].

x -1 0 1 2 3
P(X=x) 0.35 0.05 0.1 0.3 0.2

thanks for the help!
• April 13th 2010, 02:28 AM
losm1
Maybe $P(X>0.95) = 1 - P(less\ then\ 0.95) = 1 - P(-1) + P(0) = 1 - 0.4 = 0.6$ ?
• April 13th 2010, 02:35 AM
Failure
Quote:

Originally Posted by BabyMilo
Given $E[X]{\color{red}=}0.95$. Find the probability that X is greater than E[X].

x -1 0 1 2 3
P(X=x) 0.35 0.05 0.1 0.3 0.2

thanks for the help!

Well, exactly what they say, I suppose, i.e. they want to know

$P(X>E[X]){\color{red}=}P(X>0.95)=P(X \in \{1,2,3\})$
$=P(X=1)+P(X=2)+P(X=3)=0.1+0.3+0.2=0.6$.
BTW. It's a bit strange that they tell you what $E[X]$ is, because from the given values and probabilities of X one can easily determine the expected value.
• April 13th 2010, 02:37 AM
Quote:

Originally Posted by BabyMilo
Given E[X]=0.95. Find the probability that X is greater than E[X].

x -1 0 1 2 3
P(X=x) 0.35 0.05 0.1 0.3 0.2

thanks for the help!

Hi BabyMilo,

$E[X]=-1(0.35)+0(0.05)+1(0.1)+2(0.3)+3(0.2)=0.95$

All the probabilities sum to 1.

The probability that $X>E[X]$ is the probability of x being 1, 2 or 3,
as those values of x are >0.95

which is 0.1+0.3+0.2
• April 13th 2010, 02:39 AM
losm1
Quote:

Originally Posted by Failure
BTW. It's a bit strange that they tell you what $E[X]$ is, because from the given values and probabilities of X one can easily determine the expected value.

Yes, that is unusual (Surprised)