# Conditional normal probability, repost

1. May 8, 2013

### Gauss M.D.

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

The number of hours of sunshine in one week in a specific resort is assumed to follow a normal distribution with expectation 43 and standard deviation 17.

Family A will spend the first three weeks of the summer at the resort. Family B will spend the LAST two weeks of the summer at the resort.

Assuming everything is independent, what is the probability that family A will get atleast twice as much sunshine as family B?

2. Relevant equations

A = # of sunshine hours for family A
B = # of sunshine hours for family B

3. The attempt at a solution

E(A) = 3*E(X) = 129
σ(A) = sqrt(3)*σ(X) = 29.4

E(B) = 86
σ(B) = sqrt(2)*σ(X) = 24

So A ~ N(129, 29.4), B ~ N(86, 24)

If we let Z = A/B and try to find P(Z > 2), we do not have a linear combination of normally distributed random variables so I guess that's not the right way to go, assuming we want to do the normal table reading thing. So I don't know where to go from here. Help!

2. May 8, 2013

### Staff: Mentor

You could try an integration over A or B, and see if that works.

Edit: Oh, I did not see this. Follow awkward's hint, that is way easier.

Last edited: May 8, 2013
3. May 8, 2013

### awkward

Hint: What can you say about the distribution of the random variable A - 2B?

4. May 9, 2013

### Gauss M.D.

D'oh! Thank you so much!