Question

The sum mean and variance of a B.D. for $5$ trials is $1.8$, then the Binomial distribution (B.D.) is given by

• A. ${(0.2+0.8)}^5$
• B. ${(0.3+0.7)}^5$
• C. ${(0.4+0.6)}^5$
• D. None of these

Answer 1

The correct option is A ${(0.2+0.8)}^5$

mean$\left(m\right)$ for $B.D.=np$
Variance $\left(o^2\right)$ for $B.D.=npq$
$\therefore np(1+q)=\frac{18}{10,}p(1+q)=\frac{18}{50}$
$(1-q)(1+q)=\frac{18}{50}$
$\therefore q=0.8(q=-0.8$ can't possible as $0\le q\le 1)$
$\therefore q=0.8,p=0.2$
$\therefore B.D.={(0.2+.8)}^5$