Hi guys,

I'm new to this forum, and perhaps there already exist an answer to this question.

Anyway here I go, I have an assignment in statistics where a few questions got me puzzled.

The first one is:

"suppose x and y are jointly normally distributed with E(X)=1 and E(Y)=2, Var(X)=Var(Y)=1/3 and the correlation coefficient is p=1/2

Calculate P(2.2<Y<3.2|X=3)"

In my mind P(X=3) is zero in a normal distribution curve and therefore P(2.2<Y<3.2|X=3)=0, but I probably have missed something crucial???

This is a fairly straightforward problem (and fun), and the equation that you want can be found in a lot of places so I will try to help your intuition...

Suppose height and weight are both normally distributed, and that there is significant correlation between the two variables so that taller people are more likely to weigh more (positive correlation). Again, for sake of argument, lets say that E(weight)=175lb and E(height)=60in.

Now, let's say that my height is 75in, which is way above expected. Knowing my height, what is probability that I weigh between 175 and 300? Your intuition should say that since I am taller than normal, and height and weight are positively correlated, I should weigh higher than normal.

I hope this helps your intuition of the problem.

Best.

thanks hockeyfan.

I do understand the intuition but I'm not quite sure of how to calculate, been trying to find a formula everywhere.

not sure if i'm on to something but this is my latest attempt to solve the task

E(A)=E(X+Y)=E(X)+E(Y)=3

Cov(XY)=p*SD(X)*SD(Y)=1/2*1/3=1/6

Var(X+Y)=Var(X)+Var(Y)+2*Cov(XY)=2/6+2/6+2*1/6=6/6=1


P(5.2<A<6.2)=P((A1-E(A))/SD(X+Y)<Z<(A2-E(A))/SD(X+Y))=

=P((5.2-3)/(1)<Z<(6.2-3)/(1))=P(2.2<z<3.2) => P=8.21%

this doesn't feel all correct though, what have I missed?

I think I finally found an answer to my question.

E(Y)=2; E(X)=1; Var(X)=Var(Y)=1/3; p=1/2

E(Y|X=3)=bX+c

b=p*SD(Y)/SD(X)=1/2*1/3/1/3=1/2
c=E(Y)-b*E(X)=2-0.5*1=1.5

Var(Y|X=3)=(1-p^2)*Var(Y)=(1-0.5^2)*1/3=3/4*1/3=3/12=1/4
SD(Y|X=3)=1/2

E(Y|X=3)=.5*3+1.5=3

P(2.2<Y<3.2|X=3)=P((2.2-3)/1/2)<Z<(3.2-3)/1/2))=P(-1.6<Z<0.4)

==> P=0.9452-.5+0.6554-0.5=0.6006

please tell me this is correct...

I believe that is correct. Nice job!!!!!