
Probability question
I've been stuck on this question for a while...
Joe has a midterm in his 1:00pm geography class. Unfortunately, his computer class, which is scheduled to end at 12:50pm, usually runs overtime. The amount of time it takes that it ends late is normally distributed with a mean of 3 minutes and a standard deviation of 2 minutes. It then takes him an amount of time to get to his geography classroom which is normally distributed with mean 4 minutes and standard deviation 1 minute. What is the probability that Joe is late for his midterm? (Use the fact that the sum of two independent normal random variables is also normally distributed.)
Thanks (Bow)

For 2 random variables A and B:
$\displaystyle E(A+B) = E(A) + E(B)$
$\displaystyle Var(A+B) = Var(A)+Var(B)$
Which means in this case the mean time from the scheduled end of class until he can get to his test is 3+4 = 7 minutes.
The variance of this time is $\displaystyle 2^2 + 1^2 = 5$ minutes, so the standard deviation is $\displaystyle \sqrt{5} = 2.236$ minutes
You can use a calculating utility to solve the problem now, or convert to Z values and look up the probability in the standard normal distribution table.

Thank you very much for your help!