Question

The mean of a binomial distribution is 20 and the standard deviation 4. Calculate the parameters of the binomial distribution.

Answer 1

Given that mean, i.e. np = 20       ...(1)
and standard deviation, i.e. npq = 4

\[\sqrt{npq} = 4 \]

\[ \Rightarrow npq = 16 . . . (2)\]

\[\text{ Dividing eq (2) by eq (1), we get } \]

\[q = \frac{16}{20} = \frac{4}{5}\]

\[\text{ and }  p = \frac{1}{5}; \]

\[ \therefore n = \frac{Mean}{p} = 100 \]

\[P(X = r) = ^{100}{}{C}_r \left( \frac{1}{5} \right)^r \left( \frac{4}{5} \right)^{100 - r} , r = 0, 1, 2 . . . . . 100\]

\[\text{ Therefore, the parameters are n = 100 and p } = \frac{1}{5}\]