The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respective

Question

The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. Then the number of trials in the binomial distribution is

Professor PC Thomas Classes · Answer

(96) : Here, Mean = $n p$ and Variance = npq\ Now, $n p+n p q=82.5$\ $\Rightarrow \quad n p(1+q)=82.5$\ and $(n p)(n p q)=1350$\ $\Rightarrow \quad n^{2} p^{2} q=1350$\ From (i) and (ii), $\frac{q}{(1+q)^{2}}=\frac{1350}{(82.5)^{2}}=\frac{24}{121}$\ $\Rightarrow \quad 24(1+q)^{2}=121 q$\ $\Rightarrow \quad 24 q^{2}-73 q+24=0$\ $\Rightarrow q=\frac{3}{8}$ or $q=\frac{8}{3}$\ $\Rightarrow q=\frac{3}{8}$ and $p=1-\frac{3}{8}=\frac{5}{8} \quad 	herefore n=96$\