# What are the mean and standard deviation of a binomial probability distribution with n=25  and p=10/17 ?

Feb 12, 2016

$\mu = \frac{250}{17} \cong 0.588$
$\sigma = \sqrt{\frac{1750}{289}} \cong 2.46$

#### Explanation:

The following are the equations for the mean and variance of the binomial distribution (taken from Wikipedia: Binomial Distribution )

$\mu = n \cdot p$
$V a r = n \cdot p \cdot \left(1 - p\right)$

where $n$ is the number of trials, and $p$ is the probability of success in each trial. Standard Deviation is simply the square root of the variance:

$\sigma = \sqrt{V a r} = \sqrt{n \cdot p \cdot \left(1 - p\right)}$

using the values from the question

$\mu = 25 \cdot \frac{10}{17} = \frac{250}{17} \cong 0.588$
$\sigma = \sqrt{25 \cdot \frac{10}{17} \cdot \left(1 - \frac{10}{289}\right)} = \sqrt{\frac{1750}{289}} \cong 2.46$