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Question
If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.
Solution
Given: Mean = 9 and variance = 6
\[\therefore \text{ np }= 9 . . . (1) \]
\[ \text{ npq }= 6 . . . (2) \]
\[\text{ Dividing eq (2) by eq (1), we get} \]
\[ \text{ q }= \frac{2}{3}\text{ and } \text{ p = 1 - q } = \frac{1}{3}\]
\[\text{ As np = 9, substituting the value of p, we get} \]
\[\frac{\text{ n }}{3} = 9 \text{ or } \text{ n } = 27\]
\[\text{ P(X = r) } =^{27}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{27 - r} , r = 0, 1, 2 . . . . 27\]
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