Question

The difference between mean and variance of a binomial distribution is 1 and the difference of their squares is 11. Find the distribution.

Answer 1

np - npq = 1                     ...(1)

⇒   np (1 - q) = 1

⇒   npp = 1 

⇒   np² = 1

Also n²p² - n²p²q² = 11 

⇒   n²p²(1 - q²) = 11

⇒  n²p²(1 - q) ( 1 + q) = 11 

⇒  n²p² . p . (1 + q) = 11

⇒  np² np (1 + q) = 11 

⇒ np ( 1 + q) = 11

⇒ np - npq = 1         ...(1)

${\displaystyle \frac{(np+npq=11) ...\left(2\right)}{2np=12}}$

np = 6

and np² = 1

⇒ 6p = 1

⇒ p = ${\displaystyle \frac{1}{6}}$

∴ q = 1 - p = 1 - ${\displaystyle \frac{1}{6}=\frac{5}{6}}$

Now, np = 6

⇒ ${\displaystyle n\left(\frac{1}{6}\right)=6}$

⇒ n = 36

So, the binomial distribution is :

(q + p)ⁿ i.e., ${\displaystyle {(\frac{5}{6}+\frac{1}{6})}^{36}}$