Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
\[\text{ Given }: \]
\[\text{ Sum of the mean and variance } = \frac{25}{3}\]
\[ \Rightarrow \text{ np + npq } = \frac{25}{3} \]
\[ \Rightarrow \text{ np }\left( 1 + \text{ q }\right) = \frac{25}{3} . . . (1)\]
\[\text{ Product of the mean and variance } = \frac{50}{3}\]
\[ \Rightarrow \text{ np(npq) }= \frac{50}{3} . . . (2)\]
\[\text{ Dividing eq (2) by eq (1), we get } \]
\[\frac{\text{ np(npq) }}{\text{ np(1 + q) }} = \frac{50}{3} \times \frac{3}{25}\]
\[ \Rightarrow \frac{npq}{1 + q} = 2\]
\[ \Rightarrow \text{ npq }= 2(1 + q) \]
\[ \Rightarrow np(1 - p) = 2(2 - p)\]
\[ \Rightarrow np = \frac{2(2 - p)}{(1 - p)}\]
\[\text{ Substituting this value in np } + npq = \frac{25}{3}, \text{ we get } \]
\[\frac{2(2 - p)}{(1 - p)}(2 - p) = \frac{25}{3}\]
\[ \Rightarrow 6(4 - 4p + p^2 ) = 25 - 25p\]
\[ \Rightarrow 6 p^2 + p - 1 = 0\]
\[ \Rightarrow (3p - 1)(2p + 1) = 0 \]
\[ \Rightarrow p = \frac{1}{3} \text{ or } \frac{- 1}{2} . \]
\[ \text{ As p cannot be negative, the only answer for p is} \frac{1}{3} .\]
\[q = 1 - p = \frac{2}{3} \]
\[ \Rightarrow np + npq = \frac{25}{3}\]
\[ \Rightarrow n\left( \frac{1}{3} \right)\left( 1 + \frac{2}{3} \right) = \frac{25}{3}\]
\[ \Rightarrow n = 15\]
\[ \therefore P(X = r) =^{15}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{15 - r} , r = 0, 1, 2 . . . . . . 15\]
