Question

Determine the binomial distribution whose mean is 20 and variance 16.

Answer 1

Mean, i.e. np =20 ....(1)

Variance, i.e. npq =16 ....(2)

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