P(2X > 3Y - 5)

[rad6210]

Let X and Y be independent random variables, with E(X) = 1, E(Y) = 2, Var(X) = 3, and Var(Y) = 4

a) Find E(10X[sup:7qbmj2ia]2[/sup:7qbmj2ia] + 8Y[sup:7qbmj2ia]2[/sup:7qbmj2ia] - XY + 8X + 5Y - 1)
b) Assuming all variables are normally distributed, find P(2X > 3Y - 5)

I think part a is 99 (correct me if I'm wrong), but I'm not sure how to get part b ... I think you move all the variables to one side to get P(2X - 3Y > -5) , but then I don't know how to solve it. Thanks for any help!

 

[tkhunny]

1) How did you manage to miss by 4? Please demonstrate your calculation that resulted in 99.

2) E[2X-3Y] = 2E[X] - 3E[Y]

E[(2X-3Y)^2] = E[4X^2 - 12XY + 9Y^2] = 4E[X^2] - 12E[X]E[Y] + 9E{Y^2}

Var(2X-3Y) = E[(2X-3Y)^2] - E[2X-3Y]

We're almost there!

Tell me where we needed the assumption of "Independence".

 

[nannie]

I got 119 for part a.

E(X) = 1
E(Y) = 2
Var(X) = 3 = E(X^2) - [E(X)]^2 = E(X^2) -1
so E(X^2) = 4
Var(Y) = 4 = E(Y^2) - [E(Y)]^2 = E(Y^2) -4
so E(Y^2) = 8

10E(X^2) + 8E(Y^2) - E(X)E(Y) + 8E(X) + 5E(Y) - E(1) = 40 + 64 - 2 + 8 + 10 - 1 = 119
I am a student and had some trouble with this problem. I feel like I am missing something important. Please let me know if there is an error here.

I found the second part of the post very helpful. Thanks.