Normal distribution intervals.

BERRY · #1 $Old$ November 14th, 2009, 05:37 PM

An instructor has two classes, one size 30, the other size 64. From past experience, the students' mean exam score is 77 and standard deviation is 15.
What is the approximate probability that the average test score in the size 30 class is more than 1 point higher than that of the size 64 class.

This is what I have so far but don't know where to go after this:
$P(X_{30} > X_{64} + 1) = P(\frac{X_{30}-77}{15/\sqrt{30}} > \frac{X_{64}-77}{15/ \sqrt{64}} + ?)$

CaptainBlack · #2 $Old$ November 15th, 2009, 02:48 AM

Quote:

Originally Posted by BERRY $View Post$

An instructor has two classes, one size 30, the other size 64. From past experience, the students' mean exam score is 77 and standard deviation is 15.
What is the approximate probability that the average test score in the size 30 class is more than 1 point higher than that of the size 64 class.

This is what I have so far but don't know where to go after this:
$P(X_{30} > X_{64} + 1) = P(\frac{X_{30}-77}{15/\sqrt{30}} > \frac{X_{64}-77}{15/ \sqrt{64}} + ?)$

The mean class scores have means $77$ , and and variances $15^2/30$ and $15^2/64$ respectivly.

Therefore the difference in means has mean $0$ and variance $15^2/30+15^2/64$ .

Now we want to know what is the probability (assume the difference is normall distributed) that a RV $X \sim N(0, 15^2/30+15^2/64)$ is greater than $1$ .

CB