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) \[{{\left( \frac{4}{5}+\frac{5}{1} \right)}^{50}}\]
Correct Answer: B
Solution :
[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 \[={{\left( p+q \right)}^{n}}={{\left( \frac{1}{5}+\frac{4}{5} \right)}^{50}}\]
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