• # question_answer The binomial distribution whose mean is 10 and S.D. is $2\sqrt{2}$ is: A)  ${{\left( \frac{1}{5}+\frac{4}{5} \right)}^{\frac{1}{50}}}$                  B)  ${{\left( \frac{1}{5}+\frac{4}{5} \right)}^{50}}$ C)  ${{\left( \frac{4}{5}-\frac{1}{5} \right)}^{50}}$                       D)  ${{\left( \frac{4}{5}+\frac{5}{1} \right)}^{50}}$

[b] Since, Mean $=np=10$ and Standard Deviation, S.D=$\sqrt{npq}=2\sqrt{2}$ $npq=4kt2=8\,\Rightarrow 10q=8$ $\therefore q=\frac{8}{10}=\frac{4}{5}$ $\therefore p=1-q=1-\frac{4}{5}=\frac{1}{5}$ Now, $np=10$ $\therefore ~n=10\times 5=50.$ So, binomial distribution $={{\left( p+q \right)}^{n}}={{\left( \frac{1}{5}+\frac{4}{5} \right)}^{50}}$