# What are the mean and standard deviation of a binomial probability distribution with n=16  and p=12/47 ?

##### 1 Answer
Feb 22, 2016

the mean is $\frac{192}{47}$ and the standard deviation is $\frac{\sqrt{6720}}{47}$

#### Explanation:

The mean and standard deviation of the Binomial distribution are given by:

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

Note that the standard deviation is simply the square root of the variance. Plugging in the numbers we get:

$\mu = 16 \cdot \left(\frac{12}{47}\right) = \frac{192}{47} \cong 4.09$

and

$\sigma = \sqrt{\frac{192}{47} \cdot \left(1 - \frac{12}{47}\right)} = \frac{\sqrt{6720}}{47} \cong 1.74$