Question

What are the mean and standard deviation of a binomial probability distribution with `n=144` and `p=11/13`?

Answer 1

`mu = 1584/13 ~=121.8`

`sigma = 12/13sqrt(22)~=4.330`

Explanation:

The mean of a binomial distribution is simply:

`mu = n p`

Which in our case is

`mu = 144*11/13 = 1584/13 ~=121.8`

The standard deviation of the binomial distribution is given by:

`sigma=sqrt(n p (1-p))`

Which in our case is:

`sigma = sqrt(144*11/13*(2/13)) = 12/13sqrt(22)~=4.330`

A great resource for these kinds of questions is wikipedia, where they have a page for most distributions each of which has a table listing all of the properties and the associated equations:

https://en.wikipedia.org/wiki/Binomial_distribution