A random sample of size 49 is taken from a population with mean 31and standard deviation of 12.
What are the expected value and the standard deviation of the sample mean ?
Describe the probability distribution of .
What is the probability that the sample mean is greater than 32?
What is the probability that the sample mean falls between 27 and 29?
What is the probability that the sample mean will be within ±3 of the population mean?

The distribution is the normal distribution which is symmetrical about the mean. Symmetric means it looks the same on both sides of the mean.
P(sample mean > 32)
We need to find a Z-score, which shows how many standard deviations away from the mean a value is.
Z = (sample mean - population mean)/(standard deviation/square root(n))
Z = (32 - 31)/(12/square root (49))
Z = 1/(12/7)
Z = 0.58

Now find Z(0.58) on the standard normal distribution chart. Since the chart shows probabilities less than, we need to take 1- Z(0.58)
Here is the chat I used for all these. http://www.regentsprep.org/regents/math/algtrig/ats7/zchart.htm
Z(0.58) = .7190
1- .7190 = .2010
The probability it falls between 27 and 29 ... P(27 < X < 29)
We need 2 Z-scores now, one for 27 and one for 29
Z = (27 -31)/(12/sqrt(49))
Z = -4/(12/7)
Z = -2.33
and
Z = (29 - 31)/(12/sqrt(49))
Z = -2/(12/7)
Z = -1.17
Now we get Z(-1.17) and Z(-2.33) and subtract them
.1210 - .0099 = .1111
Probability sample mean is within +/- 3standard deviations we simply find Z(3) and Z(-3) and subtract
So we get .9987 - .0013 = .9974