Question

In a binomial distribution, mean is $3$ and standard deviation is $\frac{3}{2}$, then the probability function 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. ${\left(\frac{3}{4}+\frac{1}{4}\right)}^9$

Answer 1

Answer: (A)

In a binomial distribution,
Mean $=np$ , Var = npq
and standard deviation $=\sqrt{npq}$
$\therefore$ Mean $=3,S.D=\frac{3}{2}$
$\Rightarrow q=\frac{npq}{np}$ (multiply and divide by np)
$=\frac{9}{4\times 3}=\frac{3}{4}$ $\left(\because \sqrt{npq}=\frac{3}{2}\right)$
$\Rightarrow p=1-\frac{3}{4}=\frac{1}{4}(\because p=1-q)$
Now, $np=3$
$\Rightarrow n\cdot \frac{1}{4}=3$
$\Rightarrow n=12$
Therefore, binomial function is given as
$(q+p{)}^n={\left(\frac{3}{4}+\frac{1}{4}\right)}^{12}$