# What are the mean and standard deviation of a binomial probability distribution with n=18  and p=2/32 ?

Dec 23, 2015

The mean is $\mu = n \cdot p = 18 \cdot \frac{1}{16} = \frac{9}{8}$ and the standard deviation is $\sigma = \sqrt{n \cdot p \cdot \left(1 - p\right)} = \sqrt{\frac{135}{128}} = \frac{3 \sqrt{15}}{8 \sqrt{2}} = \frac{3 \sqrt{30}}{16} \approx 1.027$

#### Explanation:

A binomial random variable $X$ counts the number of successes in $n$ trials, when each trial is independent from the others with a constant probability $p$ of success.

The mean of $X$ is $\mu = n \cdot p$ and the standard deviation of $X$ is $\sigma = \sqrt{n \cdot p \cdot \left(1 - p\right)}$.