# Thread: So confused... Central Areas

1. ## So confused... Central Areas

Hey

I'm having major issues trying to figure this out...

In America, there are 2 primary standardized tests used by colleges to decide on admissions: SAT & ACT. In 2001, SAT scores had a mean of 500 & a standard deviation of 100, while ACT scores had a mean of 21.0 & a standard deviation of 4.7. In each case, the higher the mark the better the result & the greater the chance of being accepted for admission. Assume that the scores for students in general are Normally Distributed with means & standard deviations as stated above.

Question: Between what 2 scores do the central 80% of students score in the SAT tests?

Now this is what I have:

For the central 80% of SAT student test scores: pr(xL < X < xU) = 0.80
pr(xL < X < xU) = 0.80 = pr(X < xL) = 0.10 = xL =
pr(xL < X < xU) = 0.80 = pr(X < xU) = 0.90 = xU =
The central 80% of SAT test scores fall between

So I know thats the working formula, I just have no idea how to get the answer... I know we use our calculators to get the figures, but I cant remember how. Its driving me nuts, I've listened to my lectures about 6 times and gone over all my notes and this is making me draw a huge blank and I'm really confused.

Any help would be greatly appreciated!!

2. I don't know how they do it in your class but I would do:

$pr(X>\frac{x-500}{100})=0.9$

Then find the z-value for the the 90th percentile on a z-table (1.28)

So then we have:

$pr(X>\frac{x-500}{100}=1.28)$

$x=628)$

So the upper limit for the central 80% is 628

Then

$pr(X<\frac{x-500}{100})=0.1$

Then find the z-value for the the 10th percentile on a z-table (-1.28)

$pr(X>\frac{x-500}{100}=-1.28)$

$x=372$

So the upper limit for the central 80% is 372

The scores for the central 80% are between 372 and 628.