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Joint Probability Distribution + Expectations

Attachment 24893

I have this Joint Probability table: I need to figure out

a) E[X] (given Y=30)

b) E[Y] (given X=5)

I have worked something out, I'm not certain it is correct though.

For a) I did: (1 x .10) + (5 x .05) + (8 x .15) = **1.55**

and for b) : (14 x .17) + (22 x .15) + (30 x .05) + (40 x .02) + (65 x .01) = **7.91**

Can anyone tell me if I am moving along the right track?

Thanks!

Re: Joint Probability Distribution + Expectations

Quote:

Originally Posted by

**Walowizard** Attachment 24893
a) E[X] (given Y=30)

b) E[Y] (given X=5)

I have worked something out, I'm not certain it is correct though.

For a) I did: (1 x .10) + (5 x .05) + (8 x .15) =

**1.55**
and for b) : (14 x .17) + (22 x .15) + (30 x .05) + (40 x .02) + (65 x .01) =

**7.91**

I have not (will not) do the calculations.

BUT the setups are correct in both parts.

Re: Joint Probability Distribution + Expectations

Sorry, I am having one last problem with this table.

Again based on the table in my first post, I have a question that says:

**What is the variance of Y?**

I think I have to do it like this:

Var(Y) = (Probability that Y=14)*(14 - mean)^2 + P(Y=22)*(22-mean)^2 +............. P(Y=65)*(65-mean)^2

My problem is finding the Probability that Y = 14, 22, etc.... and the mean. I believe I have calculated the mean correctly though, I came up with **34.2**

If anyone can help me out it would be great :D

Re: Joint Probability Distribution + Expectations

Sorry once again for bumping this so soon, but I think I have solved it! Just want to make sure :P

So for the Variance of Y, I did this:

(.21)(14-34.2)^2 + (.23)(22-34.2)^2 + (.30)(30-34.2)^2 + (.15)(40-34.2)^2 + (.11)(65-34.2)^2 = **238.2**

Have I solved this correctly? (I am still unsure about the mean being 34.2)