Discrete Random Variables

**BabyMilo** · April 13th 2010, 01:56 AM

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

Unsure what's asking me there.

thanks for the help!

**losm1** · April 13th 2010, 02:28 AM

Maybe $P(X>0.95) = 1 - P(less\ then\ 0.95) = 1 - P(-1) + P(0) = 1 - 0.4 = 0.6$ ?

**Failure** · April 13th 2010, 02:35 AM

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

Unsure what's asking me there.

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.

**Archie Meade** · April 13th 2010, 02:37 AM

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

Unsure what's asking me there.

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

**losm1** · April 13th 2010, 02:39 AM

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

Thread: Discrete Random Variables

LinkBack

Thread Tools

Display

Discrete Random Variables

Similar Math Help Forum Discussions

Discrete random variables

Discrete random variables (2)

help with discrete random variables

What's the difference between discrete random variables and continous random variable

Discrete Random Variables

Search Tags