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If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.

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Answer Text

Solution :
consider the binomial distribution as`(1+p)^n` <br> `P(x=m) = (.^nC_mP^m)/(1+p)^n` <br> mean`= sum_m P(x=m)m= (sum .^nC_m.P^m*m)/(1+p)^n` <br> `= P/(1+p)^n*sum.^nC_m . p_(m-1)m = n*p/(1+p) = 9` <br> variance`= sum p(x=m)m^2 - (sum p(x=m)m)^2/9^2 ` <br> `6 = (sum .^nC_mP^mm^2)/(1+ p)^n - 9^2` <br> `6+9^2 = (sum.^nC_m P^m(m-1))/(1+p)^n + (sum m.^nC_m P^m)/(1+p)^n` <br> ` = p^2/(1+p)^n*n(n-1)(1+p)^(n-2) + (n*(1+p)^(n-1)p)/(1+p)^n` <br> `= (n^2-n)*p^2/(1+p)^2 + np(1+p)/(1+p)^2` <br> ` 6+81 = (n^2p^2-np^2 +np +np^2)/(1+p)^2` <br> `6 + 81 = (n^2p^2(n+np))/(np(1+p)^2)` eqn(1) <br> `(np)/(1+p) = 9` <br> `87= 81(1+np)/np` <br> `87np = 81 + 81np` <br> `26np = 81` <br> `np = 27/2` <br> `27/(2(1+p))= 9` <br> `1+p = 3/2` <br> `p=1/2` <br> `n=27` <br> `:. distribution (1+1/2)^27` <br> `P(x=m)= (.^27C_m.(1/2)^m)/(3/2)^n` <br> `= .^27C_m(1/3)^m(2/3)^(n-m)` <br> `=(1/3 + 2/3)^m` <br>

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