Question

If the mean and variance of a random variable X with a binomial distribution are 4 and 2 respectively, find P (X = 1).

Answer 1

$$\begin{gathered} \mathrm{Given},\mathit{}np=4\mathrm{and}npq=2 \\ \therefore p=1-\frac{\mathrm{Variance}}{\mathrm{Mean}} \\ =1-\frac{2}{4} \\ =\frac{1}{2} \\ \text{and}q=\frac{1}{2} \\ \text{and}n\mathit{}=\frac{np}{p} \\ =\frac{4}{{\displaystyle \frac{1}{2}}} \\ =8 \\ \mathrm{Hence},\mathrm{the}\mathrm{binomial}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P\left(X=r\right)={}^{8}C_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{8-r},r=0,1,2,3.......8 \\ \therefore P(X=1)=8{\left(\frac{1}{2}\right)}^8 \\ =\frac{1}{32} \end{gathered}$$