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

$$\begin{gathered} \sqrt{npq}=4 \\ \Rightarrow npq=16...\left(2\right) \\ \mathrm{Dividing}\mathrm{eq}\left(2\right)\mathrm{by}\mathrm{eq}\left(1\right),\mathrm{we}\mathrm{get} \\ q=\frac{16}{20}=\frac{4}{5} \\ \text{and}p=\frac{1}{5}; \\ \therefore n=\frac{\mathrm{Mean}}{p}=100 \\ P(X=r)={}^{100}C_r{\left(\frac{1}{5}\right)}^r{\left(\frac{4}{5}\right)}^{100-r},r=0,1,2.....100 \\ \mathrm{Therefore},\text{the}\mathrm{parameters}\mathrm{are}\mathit{}n=100\mathrm{and}p=\frac{1}{5} \end{gathered}$$