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

Jun 20, 2016

$\mu = \frac{1584}{13} \cong 121.8$

$\sigma = \frac{12}{13} \sqrt{22} \cong 4.330$

#### Explanation:

The mean of a binomial distribution is simply:

$\mu = n p$

Which in our case is

$\mu = 144 \cdot \frac{11}{13} = \frac{1584}{13} \cong 121.8$

The standard deviation of the binomial distribution is given by:

$\sigma = \sqrt{n p \left(1 - p\right)}$

Which in our case is:

$\sigma = \sqrt{144 \cdot \frac{11}{13} \cdot \left(\frac{2}{13}\right)} = \frac{12}{13} \sqrt{22} \cong 4.330$

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: