Question

Find 'p' of a binomial distribution for which
Mean and SD are 6 and 2 respectively.​

Answer 1

In binomial distribution:

mean = np

variance = npq

where n = number of variables or trials

p = probaility of occurence or success.

q = probability of failure

and p + q = 1

Given,

mean = np = 6

standard deviation = $\sqrt{npq}$ = 2    [variance = (standard deviation)²]

Substitute np = 6 in $\sqrt{npq}$

⇒ $\sqrt{6q}$ = 2

⇒ 6q = 4

⇒ q = 2/3

p + q = 1

so, p = 1/3

The p of given binomial distribution = 1/3.

Hope this helps you.