Mean and Standard Deviation

Date: 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/