Question

In a binomial distribution mean is $4.8$ and variance is $2.88$, then the parameter $n$ is

• A. $8$
• B. $12$
• C. $16$
• D. $20$

Answer 1

Answer: (B)

$12$

Mean $=np$
$=4.8$
Variance $=np(1-p)$
$=2.88$
Therefore
$4.8(1-p)=2.88$
$1-p=0.6$
$p=0.4$
Hence $n=\frac{4.8}{0.4}$
$=\frac{48}{4}$
$=12$
'n' being the number of trials.