# What are the mean and standard deviation of a binomial probability distribution with n=20 and p=0.9?

Jan 3, 2016

Mean $18$
Standard deviation $\frac{3}{\sqrt{5}}$

#### Explanation:

$\mu = n p$

$= \left(20\right) \cdot \left(0.9\right)$

$= 18$

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

$= \sqrt{20 \cdot \left(0.9\right) \cdot \left(0.1\right)}$

$= \frac{3}{\sqrt{5}}$

If the random variable $X$ follows the binomial distribution with parameters $n \in \mathbb{N}$ and $p \in \left[0 , 1\right]$, the probability of getting exactly $k$ successes in $n$ trials, is given by

$\text{Pr} \left(X = k\right) = n {C}_{k} \cdot {p}^{k} \cdot {\left(1 - p\right)}^{n - k}$

The mean is calculated from $\text{E} \left(X\right)$.

The standard deviation comes from the square root of the variance, "Var"(X) = "E"(X^2)-["E"(X)]^2

Where

$\text{E"(X) = sum_{k=0}^n k*"Pr} \left(X = k\right)$

$\text{E"(X^2) = sum_{k=0}^n k^2*"Pr} \left(X = k\right)$