Question

In a binomial distribution the sum and product of the mean and the variance are \[\frac{25}{3}\] and \[\frac{50}{3}\]

 respectively. Find the distribution.

 

 

Answer 1

\[\text{ Given }: \]

\[\text{ Sum of the mean and variance }  = \frac{25}{3}\]

\[ \Rightarrow \text{ np + npq } = \frac{25}{3} \]

\[ \Rightarrow \text{ np }\left( 1 + \text{ q }\right) = \frac{25}{3} . . . (1)\]

\[\text{ Product of the mean and variance } = \frac{50}{3}\]

\[ \Rightarrow \text{ np(npq) }= \frac{50}{3} . . . (2)\]

\[\text{ Dividing eq (2) by eq (1), we get } \]

\[\frac{\text{ np(npq) }}{\text{ np(1 + q) }} = \frac{50}{3} \times \frac{3}{25}\]

\[ \Rightarrow \frac{npq}{1 + q} = 2\]

\[ \Rightarrow \text{ npq }= 2(1 + q) \]

\[ \Rightarrow np(1 - p) = 2(2 - p)\]

\[ \Rightarrow np = \frac{2(2 - p)}{(1 - p)}\]

\[\text{ Substituting this value in np } + npq = \frac{25}{3}, \text{ we get } \]

\[\frac{2(2 - p)}{(1 - p)}(2 - p) = \frac{25}{3}\]

\[ \Rightarrow 6(4 - 4p + p^2 ) = 25 - 25p\]

\[ \Rightarrow 6 p^2 + p - 1 = 0\]

\[ \Rightarrow (3p - 1)(2p + 1) = 0 \]

\[ \Rightarrow p = \frac{1}{3} \text{ or }  \frac{- 1}{2} . \]

\[ \text{ As p cannot be negative, the only answer for p is} \frac{1}{3} .\]

\[q = 1 - p = \frac{2}{3} \]

\[ \Rightarrow np + npq = \frac{25}{3}\]

\[ \Rightarrow n\left( \frac{1}{3} \right)\left( 1 + \frac{2}{3} \right) = \frac{25}{3}\]

\[ \Rightarrow n = 15\]

\[ \therefore P(X = r) =^{15}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{15 - r} , r = 0, 1, 2 . . . . . . 15\]