Question

If the sum of mean and variance of $B.D$ for $5$ trials is $1.8$, the binomial distribution is

• A. $(0.8+0.2{)}^5$
• B. $(0.2+1.8{)}^5$
• C. ${\left(0.8+0.2\right)}^{10}$
• D. ${\left(0.2+1.8\right)}^{10}$

Answer 1

Answer: (A)

$(0.8+0.2{)}^5$

a binomial distribution is characterized by $2$ parameters:

$n$ - number of trials

$p$ - probability of success in one trail.

We there by know n=5, we need to find $p$ .

$Mean$ of the binomial distribution is given by,

$mean=np$

variance is given by

$variance=np(1-p)\therefore mean+variance=1.8=>5p+5p(1-p)=1.8=>5p^2-10p+1.8=0=>p=0.2or1.8$

$=>p=0.2[\because 1\ge p\ge 0]$

$q=0.8$

Binomial distribution is $(0.2+0.8{)}^n$.

Hence the answer is A