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Question
The mean of a binomial distribution is 20 and the standard deviation 4. Calculate the parameters of the binomial distribution.
Solution
Given that mean, i.e. np = 20 ...(1)
and standard deviation, i.e. npq = 4
\[\sqrt{npq} = 4 \]
\[ \Rightarrow npq = 16 . . . (2)\]
\[\text{ Dividing eq (2) by eq (1), we get } \]
\[q = \frac{16}{20} = \frac{4}{5}\]
\[\text{ and } p = \frac{1}{5}; \]
\[ \therefore n = \frac{Mean}{p} = 100 \]
\[P(X = r) = ^{100}{}{C}_r \left( \frac{1}{5} \right)^r \left( \frac{4}{5} \right)^{100 - r} , r = 0, 1, 2 . . . . . 100\]
\[\text{ Therefore, the parameters are n = 100 and p } = \frac{1}{5}\]
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