Question

The binomial distribution for which,
mean$+$$2\times$variance $=$4, mean$+$variance $=3$ is

• A. $(6,\frac{1}{3},\frac{2}{3})$
• B. $(6,\frac{1}{2},\frac{1}{2})$
• C. $(4,\frac{1}{2},\frac{1}{2})$
• D. $(4,\frac{1}{3},\frac{2}{3})$

Answer 1

Answer: (C)

$(4,\frac{1}{2},\frac{1}{2})$

Mean $=\mu =np$ and Variance $={\sigma }^2=npq$
mean$+2\times$variance $=4$, mean$+$variance $=3$
$np+2npq=4$ ; $np+npq=3$
$np+npq+npq=4$ ; $np+1=3$
$3+npq=4$ ; $np=2$
$npq=1$ ;

$2q=1$
$q=\frac{1}{2}$ and $p=1-q=\frac{1}{2}$

$np=2$
$n=\frac{2}{p}=2\times \frac{2}{1}=4$

Binomial distribution $=(4,\frac{1}{2},\frac{1}{2})$