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Question
Determine the binomial distribution whose mean is 20 and variance 16.
Solution
Mean, i.e. np =20 ....(1)
Variance, i.e. npq =16 ....(2)
\[\text{ Dividing eq (2) by eq (1), we get } \]
\[\frac{npq}{np} = \frac{16}{20}\]
\[ \Rightarrow q = \frac{4}{5}\]
\[ \Rightarrow p = 1 - q \]
\[ \therefore p = \frac{1}{5} \]
\[\text{ As np } = 20 \]
\[ \Rightarrow n = 100\]
\[ \therefore P(X = r) = ^{100}{}{C}_r \left( \frac{1}{5} \right)^r \left( \frac{4}{5} \right)^{100 - r} , r = 0, 1, 2 . . . . 100\]
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