The sum of the mean and variance of a binomial distribution is 15 and the sum of their squares is 117. The mean of the distribution is

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

$np+npq=15\text{ and }(np{)}^2+(npq{)}^2=117$     [Given]

$\therefore \frac{n^2{p}^2\left(1+q^2\right)}{(np+npq{)}^2}=\frac{117}{{15}^2}$

$$\begin{gathered} \Rightarrow \frac{1+q^2}{(1+q{)}^2}=\frac{117}{225}\Rightarrow \frac{1+q^2}{(1+q{)}^2}=\frac{13}{25} \\ 25\left(1+q^2\right)=13\left(1+q^2+2q\right)\Rightarrow 12q^2-26q+12=0 \\ \Rightarrow 6q^2-13q+6=0\Rightarrow \left(2q-3\right)\left(3q-2\right)=0\Rightarrow q=\frac{2}{3} \\ and p=1-q=1-\frac{2}{3} \end{gathered}$$

 and $p=\frac{1}{3}$

$$\begin{gathered} \therefore np+npq=15\Rightarrow n\times \frac{1}{3}+n\times \frac{2}{9}=15 \\ \Rightarrow \frac{5n}{9}=15\Rightarrow n=27 \end{gathered}$$

Hence, mean = np = 9.