Dan's Segment 10
1. This problem can be completed just by thinking it through. The mean value of a uniform(0,1) distribution is .5, so we can assume the sum of 12 random deviates drawn from such a distribution will be close to 6. Of course, subtracting 6 from 6 leaves 0, which is the most probable draw from a normal(0,1) distribution.
Just For Fun
1. I'm not sure how I would compute this analytically, but the idea is very similar to problem 1 from the To Calculate section above. The mean value of the uniform distribution is 0.5, so on average we would expect two random draws to add up to one. In order for the sum to exceed one, we would probably need three draws.