Not sure what to do with this statistical question: given that X is
normally distributed with a mean of 20 and standard deviation of 4,
how would one calculate the probability that the sample mean x exceed
25 based on a sample size of n=64 ?

If X~N(20;4) then ^X~N(20;4/sqrt(64)) = N(20;1/2)

It follow that P(^X > 25) = P((^X - 20)/(1/2) > (25-20)/(1/2))
                          = P(Z > 10),  where Z~N(0;1) 
                          = 1 (according to table)

Use the normal distribution formula

A normal distribution in a variate X with mean mu and variance sigma^2
is a statistic distribution with probability function
P(x)==1/(sigma*sqrt(2pi))e^(-(x-mu)^2/(2sigma^2))

Integrate this equation over 25 to infinity

Hope this will answer ur question

The standard deviation of the sample mean is the standard deviation of
the distribution divided by the square root of n.  So the sample mean
is normally distributed with mean 20 and standard deviation 4/sqrt(64)
= 0.5.  So you are asking how often you will get something greater
than 10 standard deviations above the mean of a normal distibution. 
The practical answer is never.  For all intents and purposes, the
probability is zero.  Maybe someone can tell how different erf(10) is
from 1.  I think the answer is somewhere in the neighborhood of
10^-30.

erf(10) = 1 - 4.84E-45

So your probability is 2.42E-45

Winning the lottery 5 times in a row is more likely.

im sorry but i can t make heads or tails out of which answer is
correct or helpful... can anyone simplify the above info?

I would go with rracecarr's two posts.

The idea is that a sample of 64 has less variance (and therefore a
smaller standard deviation) than a sample of one.  Therefore the mean
of a sample of size 64 will stay closer to the mean of the entire
population than a sample of size one.  As rracecarr noted, the
standard deviation of the sample is sqrt(64) or 8 times smaller than
the standard deviation of the distribution.  So the standard deviation
of the sample is 0.5 instead of 4.  The difference between the mean of
the population and the sample is 5, which is 5/0.5 = 10 standard
deviations.  The likelihood of something being 10 standard deviations
away from the mean is virtually zero.

ah that really clears things up ... thanks for your help!