# help

• May 29th 2006, 12:21 PM
skhan
help
Historically, final exam scores in regular session Psychology 281 have been normally distributed with a mean of 72%. 95% of all students obtain scores between 52 and 92.

What percentage of the students obtain scores between 80 and 86? [13.24%]

If a random sample of 45 Psych 281 students was selected at random, what is the probability that the sample's mean final exam score would be greater than 68? [0.9957]
• May 29th 2006, 12:26 PM
ThePerfectHacker
Quote:

Originally Posted by skhan
Historically, final exam scores in regular session Psychology 281 have been normally distributed with a mean of 72%. 95% of all students obtain scores between 52 and 92.

What percentage of the students obtain scores between 80 and 86? [13.24%]

If a random sample of 45 Psych 281 students was selected at random, what is the probability that the sample's mean final exam score would be greater than 68? [0.9957]

You need to find standard deviation first. Since 95% is two standard deviations you have,
$72+2\sigma=92$ thus, $\sigma=10$.
To find $P(80\leq x\leq 86)$ find,
$P(72\leq x\leq 86)-P(72\leq x \leq 80)$
You do this by finding the z-scores which are,
$z=1.4,.8$ respectively. Looking up at the charts we have, that 1.4 gives .4192 and .8 gives .2881 thus, subtract them to get, 13.11%
• May 29th 2006, 12:38 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
You need to find standard deviation first. Since 95% is two standard deviations

In fact 95% corresponds to about +/-1.96 standard deviations.

RonL
• May 29th 2006, 01:10 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
In fact 95% corresponds to about +/-1.96 standard deviations.

RonL

Explains why there is a small discrepency between my answers and his book's answer.