Question

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

Answer 1

Let n and p be the parameters of the distribution. Then,

Mean = np and Variance = npq

Given,

$n=5$ and, Mean + Variance = 1.8

$np+npq=1.8$

$5p+5pq=1.8$

$5p+5p(1-p)=1.8$

$5p^2-10p+1.8=0$

$p^2-2p+0.36=0$

$(p-0.2)(p-1.8)=0$

$p=0.2$

Thus,

$n=5,p=0.2,q=0.8$

Thus, if X denotes the binomial variate, then

$P(X=r)={5_C}_r\left(0.2{)}^r\right(0.8{)}^{5-r},r=0,1,2,3,4,5$.

This is the required binomial distribution.