BINOMIAL DISTRIBUTION

The binomial distribution is used when there are exactly two mutually exclusive outcomes of a trial. These outcomes are appropriately labeled "success" and "failure". The binomial distribution is used to obtain the probability of observing x successes in N trials, with the probability of success on a single trial denoted by p. The binomial distribution assumes that p is fixed for all trials.

The formula for the binomial probability mass function is

P(x,p,n) = (n x)*(p**x)*(1-p)**(n-x) for x = 0, 1, 2, ..., n

where

(n x) = n!/(x!(n-x)!

MEAN VALUE FOR THE BINOMIAL DISTRIBUTION

First, lets remember that the binomial distribution was called binomial because of its property:

                                                         (p + q )n = 1

this is true for any n value because p + q = 1.

From Algebra we remember that we can develop the binomial (p + q )n as a series with terms (By the way, I find this  a very good exercise for algebra!!) :

                   pn, pn-1 ,pn-2, pn-3 ,........p0 , where the coefficients can be calculated using the formula given before, or using the Pascal Triangle:

                              n                                          coefficients

                              1                                            1        1
                              2                                       1       2        1
                              3                                    1     3        3       1
                              4                                 1    4       6       4       1
                              5                               1    5    10     10      5     1
                              6                            1    6    15    20    15      6    1
                              ...                               .............................................

to find the mean value of the binomial distribution we will use the induction method. We must remember that the mean value for any distribution P(x) is given by the expression:

                                                

       for n = 1:

       = 1*p + 0*(1-p) = 1p

       for n=2

         = 2*p2 + 1*2* p*q + 0*1*q2 = 2p

        for n = 3

        = 3*p3 + 2*3*p2*q + 1*3* p*q2 + 0*1*q3 = 3*(p+q)2*p = 3p

      for n = 4

       = 4*p4 + 3*4*p3*q  + 2*6*p2*q2 + 1*4* p*q3 + 0*1*q4 = 4p*(p+q)3 =  4p

           ....and so on.....

                          for n = any integer                 = np

STANDARD DEVIATION FOR THE BINOMIAL DISTRIBUTION

We must remember that the variance is equal to the standard deviation squared. The variance is defined as:

 To find the standard deviation (σ) for the binomial distribution, we will follow a similar method we used to deduce the mean value. We will use the induction method:

       for n = 1:

       σ2 = (1-p)2 *p+ (0-p)2 * (1-p)= p(1-p) = pq

       for n=2

       σ2 =(2-2p)2 *p2+ (1-2p)2 *2*p*q+ (0-2p)2 * q2= 2pq[4pq + 1 - 4p + 4p2] = 2pq 

        for n = 3

       σ2 =(3-3p)2 *p+ (2-3p)2 *3*p2*q+ (1-3p)2 *3*p*q2+ (0-3p)2 * q3= 3pq 

       for n = 4

  σ2 =(4-4p)2 *p4+(3-4p)2 *4*p3*q  + (2-4p)2 *6*p2*q2+ (1-4p)2 *4*p*q3+ (0-4p)2*q4= 4pq   

              ....and so on.....

                          for n = any integer 

                      variance =  σ2  =npq,

               and the standard deviation is :

                                           

 

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