If the sum of the mean and variance of a binomial distribution for 5 trials is `1.8`; find the distribution.

Explanation

We know that
mean= np and variance =npq
If is being given that n=5 and mean + variance =1.8 ,
`:. , np+ npq =1.8, " where " n=5`
`hArr 5p +5npq = 1.8`
`hArr p+ p(1-p) =0.36 [ :' q=(1-p)]`
`hArr p^(2) -2p + 0.36 =0`
`hArr 100p^(2) -200p+36 =0`
`hArr 25p^(2) -50p+9=0`
`hArr 25p^(2) -45p -5p+9=0`
`hArr 5p(5p-9) =(5p-9) =0`
`hArr (5p-9)(5p-1)=0`
`hArr p=(1)/(5)+0.2 [ :' " p cannot exceed"1]`
Thus n= 5,p =0.2 and q=(1-p) =(1-0.2) =0.8
Let X be denote the binomial variate . Then the required distribution is
`P(X=r) =.^(n)C_(r ) .p^(r ).q^((n-r)) =.^(5)C_(r ) .(0,2)^(r ) .(0.8)^((5-r))`
where r=0 ,1,2,3,4,5