Extra Exercise 10
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Expected value applet:
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With two dice, the mean is 7, computed as follows. If X denotes the outcome
of a single die, we have
  EX = 1*(1/6) + 2*(1/6) + 3*(1/6) + 4*(1/6) + 5*(1/6) + 6*(1/6) = 21/6 = 3.5
Hence, if X1 and X2 denote the outcomes of the two dice, we have
  E(X1+X2) = EX1 + EX2 = 3.5 + 3.5 = 7.

Central limit theorem and Normal approximation applets:
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The approximation looks quite good for n=20 for the exponential
distribution, but even for n=100 a small (systematic) skewness is seen
in the sampling distribution. Here "systematic" means that it
consistently appears when we sample new distributions.

For the binomial distributions, the approximation visually appears quite good
- for n=10 for the binary distribution with p=0.5
- for n=15 for the binary distribution with p=0.7
- for n=30 for the binary distribution with p=0.9
- for n=80 for the binary distribution with p=0.95

From this list it appears that a skewness in the distribution of each
component of the sum (average) strongly affects with quality of the approximation.
For the binary distributions with p=0.9 or 0.95, it takes a large number of
observations to smooth out the contributions from the large peak at 1.