# What are the mean and standard deviation of a binomial probability distribution with n=100  and p=1/10 ?

Mar 16, 2016

$\mu = n p = 100 \cdot \frac{1}{10} = 10$
$\sigma = \sqrt{n p q} = \sqrt{100 \cdot \frac{1}{10} \cdot \frac{9}{10}} = 3$

#### Explanation:

For Binomial Probability Distribution:
Mean, $\mu = n p$ and Variance, sigma^2 = n*p*q; sigma=sqrt(npq)
Given that $p = \frac{1}{10} \implies q = \frac{9}{10}$ thus,
$\mu = n p = 100 \cdot \frac{1}{10} = 10$
$\sigma = \sqrt{n p q} = \sqrt{100 \cdot \frac{1}{10} \cdot \frac{9}{10}} = 3$