Let X represent the SAT score of an entering freshman at University X. The random variable X is shown to have a N(1200, 90) distribution. Let Y represent the SAT score of an entering freshman at University Y. The random variable Y is know to have N(1215, 110) distribution. A random sample of 100 freshman is obtained form each university. Let X bar = the sample mean of the 100 scores from university X and Y bar = the sample mean of the 100 scores from University Y. What is the probability that X bar will be greater than Y bar?

Question

valpey · Answer

Try the distribution of Y-X. The mean of Y-X is 15 and the variance is 90^2 + 110^2 so the S.D. of Y-X is:
$$\sqrt{90^2 + 110^2} = 142.13$$
The S.D. of Ybar-Xbar should be one tenth of that (14.213).
You just need the Z score of 15/14.213.

zarkon · Answer

typicall the notation is $N(\mu,\sigma^2)$

valpey · Answer

Oh, even easier (although I don't believe that the S.D. of any university's SAT scores is single digits).

zarkon · Answer

I doubt it is too...just going by what the notation usually means.

valpey · Answer

It is safe to assume in this case that we are operating under a $$N(\mu,\sigma)$$ paradigm. This is a common enough occurrence as much as I prefer to think in variance terms as well.

zarkon · Answer

That would be weird notation.  I have a lot of probability/statistics books.  All use the $N(\mu,\sigma^2)$ notation.