# Homework Help: X & Y Normal Distribution

1. Aug 13, 2013

### joemama69

1. The problem statement, all variables and given/known data

X refers to score distribution in Math and Y refers to score distribution in Stat in a certain degree course exam. It is known that X~N(mean = 62, sigma=7) while Y~N(mean = 68, sigma=10). If X and Y are independent, find (i) P[X+Y>120]; (ii) P[X<Y]; (iii) P[X+Y>140]; (iv)P[100 < X+Y < 150].

2. Relevant equations

3. The attempt at a solution

P(X+Y>120)...

So I have to combine X & Y into one Norm Distribution by find there combined E(x) & V(x)

E(X+Y)=E(X)+E(Y)=62+68=130

V(X+Y)=V(X)+V(Y)=7^2 + 10^2 = 149, Standard Dev = sqrt(149)

So where Z = X + Y > 120, Z~N(130,sqrt(149))

P(X+Y>120) = P(Z>120)=1-PHI((130-120)/sqrt(149))

Is this correct so far....

P(X<Y)=P(X-Y<0)

E(X-Y)=E(X)-E(Y) = 62-68=-6

V(X-Y) = V(X)+V(Y)=sqrt(149)

W~N(-6,sqrt(149))

P(X-Y<0)=P(W<0)=PHI((-6-0)\sqrt(149))

How does this look. The rest are basically the same I think if I got these two conceptualy correct. any issues?

2. Aug 13, 2013

I haven't checked down to the arithmetic level but the approach looks good. One thing that helped me keep things straight notationally was this: When dealing with the first part, where you are working with the sum of two random variables, let $S = X + Y$. Then the work you have shows that $S$ has a normal distribution with a particular mean and standard deviation, and I just found it easier to write
$P(S > 120)$ instead of the way you had it. Similar for the difference.

After a final glance I do see one item I believe you should reconsider. Do you REALLY want to write
$$V(X-Y) = V(X) + V(Y) = sqrt(149)$$

(I am pointing my comment at what you have after the second equal sign) - remember in those steps you are finding a variance.

Last edited by a moderator: Aug 14, 2013
3. Aug 13, 2013

### Ray Vickson

The equation $V(X-Y) = V(X) + V(y)$ is true if X and Y are independent, so what he wrote is correct. In fact, if $a$ and $b$ are constants, we have
$$V(aX + bY) = a^2 V(X) + b^2 V(Y)$$ for independent (or uncorrelated) X and Y, whether normally-distributed or not. It is a general result.

4. Aug 14, 2013

I am aware of all this. You missed my point, however: it is likely my comment wasn't clear. The OP is attempting to calculate the variance of X - Y: the written equation was (my comments added)
$$V(X-Y) = \underbrace{V(X) + V(Y)}_{\text{Okay here}} = \overbrace{\text{sqrt}(149)}^{\text{My concern}}$$

For emphasis, compare the ending of the above work to the ending of the line in which the variance of $V(X + Y)$ was correctly calculated.

Marginally related comment: apologies for missing the fact that I screwed up a tag in one of my earlier posts.

5. Aug 14, 2013

### Ray Vickson

Just to be clear: you (correctly) want him to write '149' instead of 'sqrt(149)'. To my shame, I missed that the first time around!

6. Aug 14, 2013

"Just to be clear: you (correctly) want him to write '149' instead of 'sqrt(149)'. To my shame, I missed that the first time around!"

No big deal. As I noted, I was not entirely clear in my original note. That comes, I am afraid, from reviewing student papers and wanting to point out issues for them without giving the whole thing away.

7. Aug 14, 2013

### Ray Vickson

I agree with the 'not giving away' part, but he did already have the correct answer---or at least, the correct numerical inputs---so I did not feel guilty about laying out some more detail.

8. Aug 14, 2013