Q. In a binomial distribution, mean is 3 and standard deviation is $\frac{3}{2}$, then the probability distribution is
A) $\left( \frac{3}{4} + \frac{1}{4} \right)^{12}$
B) $\left( \frac{1}{4} + \frac{3}{4} \right)^{12}$
C) $\left( \frac{1}{4} + \frac{3}{4} \right)^{9}$
D) $\left( \frac{3}{4} + \frac{1}{4} \right)^{9}$
Explanation
(a)
Given, $np = 3 = \sqrt{npq} = \frac{3}{2}$
$\Rightarrow \ q = \frac{npq}{np} = \frac{9}{4 \times 3} = \frac{3}{4}$
$\Rightarrow p = 1 - \frac{3}{4} = \frac{1}{4}$
Also, $np = 3\ \Rightarrow \ n = 12$
Hence, binomial distribution is
$(q + p)^{n} = \left( \frac{3}{4} + \frac{1}{4} \right)^{12}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.