Question

The binomial distribution whose mean is $9$ and S.D. is $\frac{3}{2}$ is

• A. ${(\frac{1}{2}+\frac{3}{4})}^{12}$
• B. ${(\frac{1}{3}+\frac{2}{3})}^{12}$
• C. ${(\frac{1}{4}+\frac{3}{4})}^{11}$
• D. ${(\frac{3}{4}+\frac{1}{4})}^{12}$

Answer 1

The correct option is D ${(\frac{3}{4}+\frac{1}{4})}^{12}$

Let X be the binomial distribution

X=B(n,p) =($p+q{)}^n$

$p+q=1$

Mean E(X)=np=9

and standard deviation $=\sqrt{npq}=\frac{3}{2}-\left(1\right)$

therefore, Variance V(X) $=npq=\frac{9}{4}-\left(2\right)$

Divide equation(2) by equation(1)

$\frac{npq}{np}=\frac{9}{4\times 9}⟹q=\frac{1}{4}$

so $p=1-q=1-\frac{1}{4}=\frac{3}{4}$

Therefore, $n=\frac{9}{p}=\frac{9\times 4}{3}=12$

We get X=B($12,\frac{3}{4})$

The required Binomal distribution is

X=$(p+q{)}^n=(\frac{3}{4}+\frac{1}{4}{)}^{12}$