Question

If the standard deviation of the binomial distribution $(q+p{)}^{16}$ is 2, then mean is

• A. $2$
• B. $4$
• C. $8$
• D. $7$

Answer 1

Answer: (C)

$8$

$S.D^2=Variance=np(1-p)$
$=2^2=4$
Hence $np(1-p)=4$
No. of trials is $16$.
Therefore $n=16$.
Hence $16p(1-p)=4$
$p(1-p)=\frac{1}{4}$

$p(1-p)=\frac{1}{2}(1-\frac{1}{2})$

Comparing, we get $p=\frac{1}{2}$

Therefore

$Mean$$=np$
$=\frac{16}{2}$

$=8$.