Example with Simulated Data Sets

Imagine rolling a fair 6-sided die many times and recording the results.

Mean of the parent distribution:

meanParent = Sum[i/6,{i,1,6}]

varianceParent = Sum[(i-meanParent)^2,{i,1,6}]/6

sdParent = Sqrt[varianceParent]//N

One roll of the die :

RandomInteger[{1,6}]

Multiple rolls :

nRolls = 100;
simData =RandomInteger[{1,6},nRolls]

Histogram[simData]

Mean[simData]//N

StandardDeviation[simData]//N

Average of multiple rolls :

Mean[RandomInteger[{1,6},nRolls]]

OR :

Total[RandomInteger[{1,6},nRolls]]/nRolls//N

Data for multiple "experiments" to get distribution of sample means. (Nest previous highlghted command in a Table[ ] function.)

nEx = 200;
meanDistribution =Table[Mean[RandomInteger[{1,6},nRolls]],{nEx}]//N

Histogram[meanDistribution,PlotRange->{{1,6},All}]

Mean[meanDistribution]

StandardDeviation[meanDistribution]

Closer look at distribution of the means  (don’t force display over full 0⟶6 range):

Histogram[meanDistribution]

Distribution looks Gaussian.

Central Limit Theorem

In the notation of Hughes and Hase on p. 33 (at the end of the introduction to Section 3.5),
the values x1, x2, x3, etc., are the results of individual rolls of the die, and N corresponds the number of rolls that are going to be averaged (nRolls) to get a sample mean.  The number of “experiments” (nEx) is the number of times we calculate the mean to get the distribution of the sample means. 
Repeating from above:

Mean[meanDistribution]

StandardDeviation[meanDistribution]

The standard deviaion of the distribution of the mean should be σParent/Sqrt[N] :

sdParent/Sqrt[nRolls]

Standard distribution appears consistent with Central Limit Theorem.