Mean and Standard DeviationDate: 03/01/99 at 19:08:48 From: Maria Alejandra Martino Subject: Mean and Standard Deviation I have to prove that the formula for the mean and standard deviation for a binomial distribution are: u = np and s = the square root of (npq), where n is the total number of trials, p is the probability of success, and q is the probability of failure, u is the mean and s is the standard deviation. I know that u = summation of x*P(X), where P(x) is the probability of x. I also have that P(x) = (combination of n and x)*[(p)^(x)]*[(q)^(n- x)] and that p + q = 1 (or, q = 1 - p) so that u = Summation of {x*(combination of n and x)*[(p)^(x)]*[(q)^(n-x)]} from this part on I do not know what to do. Please help me. Date: 03/02/99 at 10:14:34 From: Doctor Anthony Subject: Re: Mean and Standard Deviation One of the easiest proofs is as follows: In any one trial the number of successes = 1 with probability p " " " = 0 " q So if x1 = number of successes in trial 1 x2 = " " " 2 ........................................ xn = " " " n If X = number of successes in n trials then X = x1 + x2 + ....... + xn E(X) = E(x1) + E(x2) + ..... + E(xn) but E(x1) = 1.p + 0.q = p similarly E(x2) = E(x3) = .... = p Therefore E(X) = p + p + p + ...... + p = np Var(x1) = E(x1^2) - p^2 and E(x1^2) = 1^2.p + 0^2.q = p and so Var(x1) = p - p^2 = p(1-p) = pq Var(X) = Var(x1 + x2 + x3 + ..... + xn) and since the x's are independent = Var(x1) + var(x2) + .... + Var(xn) = pq + pq + pq + ..... + pq = npq - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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