# What are the mean and standard deviation of a binomial probability distribution with n=190  and p=7/11 ?

mean $\mu = 120 \frac{10}{11} = 120.909090$
standard deviation $\sigma = 6.630757283$

#### Explanation:

For Binomial Probability Distribution
mean $\mu = n p$ and
standard deviation $\sigma = \sqrt{n p q}$

Given $n = 190$ and $p = \frac{7}{11}$

Compute the mean:
$\mu = n p = 190 \cdot \frac{7}{11} = \frac{1330}{11} = 120 \frac{10}{11} = 120.909090$

Compute the standard deviation $\sigma$

$\sigma = \sqrt{n p \left(1 - p\right)} = \sqrt{190 \cdot \frac{7}{11} \cdot \left(1 - \frac{7}{11}\right)}$

$\sigma = \sqrt{190 \cdot \frac{7}{11} \cdot \left(\frac{4}{11}\right)} = \frac{2 \sqrt{1330}}{11}$

$\sigma = 6.630757283$

God bless...I hope the explanation is useful.