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 where 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 µ of the binomial distribution we will use the induction method. We must remember that the mean value µ for any distribution P(x) is given by the expression: for n = 1:        µ = 1*p + 0*(1-p) = 1p        for n=2         µ = 2*p2 + 1*2* p*q + 0*1*q2 = 2p         for n = 3        µ = 3*p3 + 2*3*p2*q + 1*3* p*q2 + 0*1*q3 = 3*(p+q)2*p = 3p       for n = 4       µ = 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                 µ = 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 *p3  + (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 : BACK

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