If the mean of a binomial distribution is 81, then find the interval in which standard deviation of the binomial distribution lies.

$[0,9)$

The correct answer is Option (3) → $[0,9)$

Given mean = $np = 81$

∴ Standard deviation = $\sqrt{npq} = \sqrt{81q}=9\sqrt{q}$

$⇒ S.D. <9$

Also $S.D. ≥0$

$⇒ 0≤S.D. <9$

∴  Standard deviation lies in the interval $[0, 9)$.