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Let X and Y be indep. standard normal variables. Find:

a) P(3X + 2Y > 5): This is just 1 - phi(5/sqrt(13)) from the fact that the mean is 0, and the std. deviation is sqrt(13). DId you get 0.0838 as the probability for this one?

b) P(Min(X,Y) < 1): This is equal to P(X < 1 or Y <1) = 2P(X<1) - P(X^Y < 1) = 2(0.84) - (0.84)^2, right?

c) P(|min(X,Y)| < 1) = P(-1 < min(X,Y) < 1) = 2*0.68^2 - (0.68)^2 ?

d) P(Min(X,Y) > max(X,Y) - 1)

Did you get the answer to be 0.7794?

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You've asked 12 other questions, all of which have been answered (except for the one that was closed), and you haven't accepted any of the answers. Is there a reason for this? –  joriki Apr 15 '11 at 8:56
    
You would have done better putting this on your similar earlier question as an edit to the question or as a comment on the reply –  Henry Apr 15 '11 at 10:28
    
I don't know how to accept answers. What can I click? I tried putting a green check mark, but somehow that doesn't work –  mary Apr 16 '11 at 0:38

1 Answer 1

up vote 1 down vote accepted

(b) Looks essentially right. The notation is wrong. In your expression you probably intended something like $2P(1) -P((X<1)\cap(Y<1))$. It would have been conceptually a little simpler to use $1-P(X>1)P(Y>1)$.

(c) The answer looks right. A careful grader would consider it insufficiently explained.

(d) Here there is no description of what you tried. My approach would be as follows. Let $W=X-Y$. Then $W=(1)X+(-1)Y$, so $W$ is of the form $aX+bY$ where $X$ and $Y$ are standard normal. Thus $X-Y$ is normal with mean $0$ and variance $(1)^2(1)+(-1)^2(1)$, which is $2$. As I had pointed out in the hint, you need the probability that $|W|<1$.

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If $X,Y$ have mean 0, the mean of $3X+2Y$ is 0, not 5. –  Nate Eldredge Apr 15 '11 at 14:52
    
@Nate Eldredge: Thanks! –  André Nicolas Apr 15 '11 at 17:03
    
@user6312 You might care to correct your other answer, the one on the page math.stackexchange.com/questions/32935. –  Did Apr 19 '11 at 7:44

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