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

Solution:

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

$n p + n p q = 15 and (n p)^{2} + (n p q)^{2} = 117$ [Given]

$∴ \frac{n^{2} p^{2} (1 + q^{2})}{(n p + n p q)^{2}} = \frac{117}{15^{2}}$

$\Rightarrow \frac{1 + q^{2}}{(1 + q)^{2}} = \frac{117}{225} \Rightarrow \frac{1 + q^{2}}{(1 + q)^{2}} = \frac{13}{25} 25 (1 + q^{2}) = 13 (1 + q^{2} + 2 q) \Rightarrow 12 q^{2} - 26 q + 12 = 0 \Rightarrow 6 q^{2} - 13 q + 6 = 0 \Rightarrow (2 q - 3) (3 q - 2) = 0 \Rightarrow q = \frac{2}{3} a n d p = 1 - q = 1 - \frac{2}{3}$

and $p = \frac{1}{3}$

$∴ n p + n p q = 15 \Rightarrow n \times \frac{1}{3} + n \times \frac{2}{9} = 15 \Rightarrow \frac{5 n}{9} = 15 \Rightarrow n = 27$

Hence, mean = np = 9.

Solution:

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