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

a) $P(3X + 2Y \gt 5)$: This is just $1 - \phi(\frac{5}{\sqrt{13}})$ 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) $P(Min(X,Y) \lt 1)$: This is equal to $P(X \lt 1 or Y \lt 1) = 2P(X\lt 1) - P(X^Y \lt 1) = 2(0.84) - (0.84)^2$, right?

c) $P(\mid min(X,Y)\mid < 1) = P(-1 \lt min(X,Y) \lt 1) = 2*0.68^2 - (0.68)^2$ ?

d) $P(Min(X,Y) \gt max(X,Y) - 1)$

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
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 – Did Apr 19 '11 at 7:44

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