Question

Determine the binomial distribution whose mean is 9 and variance 9/4.

Answer 1

It is given that mean, i.e. np = 9 and variance, i.e. npq =$\frac{9}{4}$

$$\begin{gathered} \therefore \frac{npq}{np}=\frac{9}{4}\times \frac{1}{9}=\frac{1}{4} \\ \mathrm{Hence},q=\frac{1}{4} \\ \text{and}p\mathit{}=1-q=\frac{3}{4} \\ \mathrm{When}p=\frac{3}{4}, \\ np\mathit{}=\frac{3n}{4} \\ 9=\frac{3n}{4} \\ \Rightarrow n=12 \\ P(X=r)={}^{12}C_r{\left(\frac{3}{4}\right)}^r{\left(\frac{1}{4}\right)}^{12-r},r=0,1,2,3,4,5,.....12 \end{gathered}$$