Question

Find n and p for a binomial distribution if mean = 6 and standard deviation = 2. Also find p (x=1).

Answer 1

It is stated that the mean $=6$ and the standard deviation $=2$.

Therefore,

$np=6.....\left(1\right)$

$npq={\left(2\right)}^2$

$\Rightarrow np(1-p)=4[\because p+q=1]$

$\Rightarrow 6(1-p)=4\left[\text{From}\left(i\right)\right]$

$\Rightarrow 1-p=\frac{4}{6}$

$\Rightarrow p=1-\frac{2}{3}=\frac{1}{3}$

Substituting the value of $p$ in equation $\left(1\right)$, we have

$n\times \frac{1}{3}=6$

$\Rightarrow n=6\times 3=18$

$\because p+q=1$

$\Rightarrow q=1-p=\frac{2}{3}$

Now,

$P(x=1)={}^{18}{C}_1{\left(\frac{1}{3}\right)}^1{\left(\frac{2}{3}\right)}^{17}=\frac{{\left(2\right)}^{18}}{{\left(3\right)}^{16}}$