Variance and Standard Deviation

In[64]:=

Var = Underoverscript[∑, x = lowsupport, arg3] (x - ExpValue)^2 pmf[x, n, p]     

Out[64]=

2.88

In[65]:=

StdDev = N[Var^(1/2)]

Out[65]=

1.69706

As with the expected value, it can be shown that the variance of a Binomial random variable is np(1-p)

In[66]:=

n * p * (1 - p)

Out[66]=

2.88

In[67]:=

End[(*BinomialProperties*)]

Out[67]=

BinomialProperties`

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