~Finding the mean (m), variance (s²), and standard deviation (s) of a Binomial Distribution.

~We need to know n (number of trials), p (probability of a success on a given trial).

~Note:  q = 1 - p (probability of a failure on a given trial).

~Recall the formulas for the mean & variance of a Probability Distribution: (lesson #10)

(1)  m = ∑xp(x)    and     s² =∑x²p(x) - m², where x represents different outcomes in the sample space and p(x) their probabilities. Remember,
∑p(x) = 1.

~A binomial distribution is just a special case with a two outcome sample space. i.e.,(head, tail), (male, female), (win, lose), (yes, no), & so on.  (one is a success & the other a failure---depends on the focus of our calculation).

~Let x = 1 be a success & x = 0 be a failure. Then p(x) = p for a success & q for a failure. Substituting these values into the equations in (1), we get the following:

m = (1)(p) + (0)(q) = p (for one trial).   For n trials, we have  m = n p (for a binomial distribution).

s²=(1)²p+(0)²q-p²=p-p²=p(1-p)=pq (for one trial). For n trials, we have s²=npq or  s = sqr(npq)

~Note:  The above is not a general proof, just one way of getting the formulas. The actual proof is more complicated & tricky. It involves substituting the general binomial probability formula into (1).

~Ex:  Would it be unusual to get 61 heads in 100 tosses of a fair coin?
Ans: Yes, 61 is more than 2 standard deviations to the right of the mean.

~Ex:  Would it be unusual to lose 100 consecutive times when betting on a specific number on roulette?
Ans: No, 0 successes is within 2 standard deviations of the mean.

~Ex:  The M&M company claims that 10% of all of their candies are blue. To test this claim, you randomly select 100 M&M candies from a large bag & note that 5 are blue. Would you say that their claim is false based on your sample?
Ans: No, 5 is within 2 standard deviations of the mean.