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)

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