# Thread: Stats question (random variables)

1. ## Stats question (random variables)

Suppose X and Y are independent random variables with X ~ N(6,2square) and Y ~ N(4,4square) Put D = Y -X

(i) What is the probability distribution of D?
(ii) Using you answer to (i) above or otherwise, evaluate P(Y<X)

My answer for (i) D ~ N(-2,12)

I am not sure if its correct, so please help!

2. I'm not sure what you mean by 2 squared

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

Then in part 2 use D=Y-X, so you want P(D>0)
which you compute by standardizing D and using your N(0,1) tables.

3. I am not really sure how to put the little 2 here.

What i meant is two square and four square.
So basically, it is N(6,4) and N(4,16)

4. Most people write N(mean, variance)
some write N(mean, st deviation).

Your variance is 4+16=20

$P(D>0)=P\biggl({D-(-2)\over \sqrt{20}}>{0+2\over \sqrt{20}}\biggr)$

$=P\biggl(Z>{2\over \sqrt{20}}\biggr)$

5. Hi thanks for the reply. Just to confirm, D should be N(10,20) then?

I still don't really get how to do the second part.

6. no, your mean was right the first time, -2 not 10.