The heights, in inches of the five starting players, whom we will call A, B, C, D, and E, on a men’s basketball team are displayed in the following table. Here, the population of interest consists of the five players and the variable under consideration is height.
Player  A  B  C  D  E 
Height  76  78  79  81  86 
For part a, here are the samples of size 2 and their means:

































Notice that each mean is slightly different from the others. If we calculate the mean of the population, we find that the mean height for the entire population is 80. So, we also see that each sample gives an estimate that is slightly different from the population mean.
We could calculate the probability that the sample mean will exactly equal the population mean by counting the number of times that it equals the population mean and dividing by the total number of samples. This will tell us that there is a 1/10 probability that the sample mean will exactly equal the population mean.
For part (b) of the question, if we count up the number of sample means that are within 1 inch of the population mean, we find that 3 out of 10 are within 1 inch. Therefore, there is a 3/10 probability that the sample mean will be within 1 inch of the population mean.
FACT: For any sample size, the mean of all possible sample means equals the population mean.
This fact can be illustrated numerically by calculating the average of our 10 sample means and noticing that it is exactly equal to 80 (the population mean).
FACT: The standard deviation of all possible sample means is equal to sigma/sqrt(n).
So, we see that the mean of SRS of size n drawn from a large population:
FACT: Suppose that we have a SRS of size n from a population that is NORMALLY distributed with mean m and standard deviation sigma . Then, xbar, the sample mean, will be normally distributed with mean mu and standard deviation sigma/sqrt(n).
What if the population is NOT normal?
CENTRAL LIMIT THEOREM (CLT):
Suppose that we have a SRS of size n from ANY population with mean mu
and standard deviation sigma . Then, for large samples (n>=30),
xbar, the sample mean, will be approximately normally distributed
with mean mu and standard deviation sigma/sqrt(n).
Examples:
Scores on an ACT test have a normal distribution with a mean of 18.6 and a standard deviation of 5.9.
Use the distribution of the POPULATION to answer this question. This question deals with a SINGLE observation from the POPULATION.
P(X>=21) = (standardize and change to zscore) P[Z>=(21 – 18.6)/5.9]
=
P(Z>= 0.41) = 1 – P(Z<=0.41) = 1 – 0.6591 = 0.3409
P(xbar>=21) = (standardize) P[Z>= (21 – 18.6)/0.835] = P(Z>=2.87)
= 1P(Z<=2.87) = 10.9979 = 0.0021
The flaws per square yard in carpet are not normally distributed but have a mean of 1.6 and a standard deviation of 1.2. A sample of 200 square yards of carpet was taken. What is the probability that the mean number of flaws per square yard exceeds 2?
By the CLT, the sampling distribution of the sample mean is normal with mean = 1.6 and standard deviation = 1.2 / sqrt(200)=0.085 (since n>=30)
P(xbar>=2)= (standardize) P[Z >= (2 – 1.6)/0.085] = P(Z >= 4.71) = 1 – P(Z<=4.71) = 11 = 0