Question

Question #137b9

Answer 1

Mean shown here; variance is similar.

Explanation:

We begin with the mean (expected value) of a discrete distribution.
`mu = Sigma[x_iP(x_i)]`.
That is, the mean is the sum of value x probability for each outcome.

For a binomial distribution, there are only two outcomes: success and failure. Let p be the probability of success. Let n be the number of trials.

Let `nCr` be "n choose r," or the Binomial Coefficient,
`nCr = (n!)/(r!(n-r)!)`.

From the probability density function for the Binomial Distribution, we have `P(r) = nCrp^r(1-p)^(n - r)`, defined for r = 0 to n -- indicating no successes, 1 success, etc., up to n successes out of n trials.

For the Binomial Distribution,
`mu = Sigma[rP(r)]`
Note that the sum is from r = 1 to n, since the value is zero if r = 0.
`= Sigma[r*nCrp^r(1-p)^(n - r)]`
`= Sigma[r* (n!)/(r!(n-r)!)*p^r(1-p)^(n - r)]`
`= Sigma[ (n!)/((r-1)!(n-r)!)*p^r(1-p)^(n - r)]`
`= nSigma[ ((n-1)!)/((r-1)!(n-r)!)*p^r(1-p)^(n - r)]`
`= nSigma[ ((n-1)C(r-1))*p^r(1-p)^(n - r)]`
`= npSigma[ ((n-1)C(r-1))*p^(r-1)(1-p)^(n - r)]`
Regarding the sum, we are summing the binomial over all possible outcomes. Since the binomial distribution is a probability distribution, the sum over all outcomes is 1.
Therefore,
`mu = np`