# What are the mean and standard deviation of a binomial probability distribution with n=12  and p=31/32 ?

Feb 18, 2016

Mean is $11.625$ and Standard Deviation = $0.603$

#### Explanation:

The mean and standard deviation of a binomial probability distribution is given by $n p$ and $\sqrt{n p q}$, where $q - 1 - p$.

With $n = 12$ and $p = \frac{31}{32}$, $q = 1 - \frac{31}{32}$ or $q = \frac{1}{32}$,

Mean = $12 \cdot \frac{31}{32}$ = $11.625$ and

Standard Deviation = sqrt(12.(31/32)*(1/32) = $\left(\frac{1}{32}\right) \cdot \sqrt{372}$ = $\frac{19.2783}{32}$ = $0.603$.