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Question
Determine the binomial distribution whose mean is 9 and variance 9/4.
Solution
It is given that mean, i.e. np = 9 and variance, i.e. npq = \[\frac{9}{4}\]
\[\therefore \frac{\text{ npq }}{\text{ np }} = \frac{9}{4} \times \frac{1}{9} = \frac{1}{4} \]
\[\text{ Hence} , \text{ q }= \frac{1}{4} \]
\[ \text{ and } \text{ p }= \text{ 1 - q }= \frac{3}{4}\]
\[\text{ When } \text{ p }= \frac{3}{4}, \]
\[\text{ np }= \frac{\text{ 3n }}{4}\]
\[ 9 = \frac{\text{ 3n }}{4}\]
\[ \Rightarrow \text{ n } = 12\]
\[\text{ P }(\text{ X = r }) = ^{12}{}{C}_r \left( \frac{3}{4} \right)^r \left( \frac{1}{4} \right)^{12 - r} , r = 0, 1, 2, 3, 4, 5, . . . . . 12\]
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