# What are the variance and standard deviation of a binomial distribution with N=124 and p=0.85?

May 3, 2016

The variance is ${\sigma}^{2} = 15.81$ and the standard deviation is $\sigma \approx 3.98$.

#### Explanation:

In a binomial distribution we have quite nice formulas for the mean and wariance:
$\mu = N p \setminus \setminus \setminus \setminus \setminus \setminus \textrm{r} \mathmr{and} \setminus \setminus \setminus \setminus \setminus {\sigma}^{2} = N p \left(1 - p\right)$

So, the variance is ${\sigma}^{2} = N p \left(1 - p\right) = 124 \cdot 0.85 \cdot 0.15 = 15.81$.
The standard deviation is (as usual) the square root of the variance:
$\sigma = \sqrt{{\sigma}^{2}} = \sqrt{15.81} \approx 3.98$.