a) Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
b) Standard Error of the mean, SEm = SD/√(n-1)
Z = (score-mean)/SEm
Use same table.
It is well known that the heights of individual American men are normally distributed with mean 70 inches and standard deviation 2.8 inches.
The Central Limit Theorem states that if n men are randomly chosen, then their average height will also be normally distributed with mean 70 inches (so the mean is unchanged), but the standard deviation will not be 2.8 inches--it will be 2.8 divided by the square root of n (the standard deviation is smaller for groups than for individuals). This means that there is less variation among group averages than there is between individuals--I hope that makes intuitive sense.
a) How likely is it that a randomly chosen man would be more than six feet tall (i.e., what percentage of men are over six feet)?
b) How likely is it that a randomly chosen group of ten men would have an average height exceeding six feet?
Remember: The answer is never sufficient:
What's of interest is always the explanation!
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