If X is a binomial with mean 6 and variance 2, then the value of $\mathrm{P}\left(5\le \mathrm{X}\le 7\right)$ is 

Given X follows Binomial distribution 

Given that np=6

                 npq=2

Now $\frac{\mathrm{npq}}{\mathrm{np}}=\frac{2}{6} \Rightarrow \mathrm{q}=\frac{1}{3}$

Now  $\begin{array}{c}p=1-\mathrm{q}\\ =1-\frac{1}{3}\\ =\frac{2}{3}\end{array} \overline{) \begin{array}{c}\mathrm{and} \mathrm{np}=6\\ \Rightarrow \mathrm{n}\left(\frac{2}{3}\right)=6\\ \Rightarrow \mathrm{n}=9\end{array}}$           

Now  $\mathrm{P}(5\le X\le 7)=\mathrm{P}(X=5)+\mathrm{P}(X=6)+\mathrm{P}(X=7)$

         $$\begin{gathered} ={}^9{\mathrm{c}}_5 {\left(\frac{2}{3}\right)}^5 {\left(\frac{1}{3}\right)}^4+{}^9{\mathrm{c}}_6 {\left(\frac{2}{3}\right)}^6 {\left(\frac{1}{3}\right)}^3+{}^9{\mathrm{c}}_9 {\left(\frac{2}{3}\right)}^7 {\left(\frac{1}{3}\right)}^2 \\ ={}^9{\mathrm{c}}_4 {\left(\frac{2}{3}\right)}^5 {\left(\frac{1}{3}\right)}^4+{}^9{\mathrm{c}}_3 {\left(\frac{2}{3}\right)}^6 {\left(\frac{1}{3}\right)}^3+{}^9{\mathrm{c}}_2 {\left(\frac{2}{3}\right)}^7 {\left(\frac{1}{3}\right)}^2 \\ ={\left(\frac{2}{3}\right)}^5 {\left(\frac{1}{3}\right)}^2 \left[{}^9{\mathrm{c}}_4 {\left(\frac{1}{3}\right)}^2+{}^9{\mathrm{c}}_3 \left(\frac{2}{3}\right) \left(\frac{1}{3}\right)+{}^9{\mathrm{c}}_2 {\left(\frac{2}{3}\right)}^2\right] \\ =\frac{32}{243} \left(\frac{1}{9}\right) \left[\frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}\times \frac{1}{9}\times \frac{9\times 8\times 7}{3\times 2\times 1} \left(\frac{2}{9}\right)+\frac{9\times 8}{2}\times \frac{4}{9}\right] \\ =\frac{32}{2187} \left[14+\frac{56}{3}+16\right] \\ =\frac{32}{2187} \left[30+\frac{56}{3}\right]=\frac{32}{2187}\times \frac{146}{3} \\ =\frac{4672}{6561} \end{gathered}$$