Question

Find the binomial distribution when the sum of its mean and variance for five trials is $4.8$.

Answer 1

Given for a binomial distribution, the sum of its mean and variance for $5$ trials is $4.8$.

Let the parameters be $n,P$

So, $np+npq=4.8$ where $q=1-p$

$\Rightarrow 5p(1+q)=4.8$

$\Rightarrow 5p(2-p)=4.8$

$\Rightarrow 5P^2-10p+4.8=0$

$\Rightarrow 5P^2-6p-4p+4.8=0$

$\Rightarrow 5p(p-1.2)-4(p-1.2)=0$

$\Rightarrow p=\frac{4}{5}$ or $P=1.2\to xp\le 1$

So, $q=1-\frac{4}{5}=\frac{1}{5}$

$\therefore$ The binomial distribution is ,

$P(x=r)={}^5{C}_r{\left(\frac{4}{5}\right)}^r{\left(\frac{1}{5}\right)}^{5-r}$

Where $r=0,1,\mathrm{2....5}$