# What are the mean and standard deviation of a binomial probability distribution with n=2  and p=4/17 ?

The mean is $\mu = n \cdot p = 2 \cdot \frac{4}{17} = \frac{8}{17} \approx 0.4706$ and the standard deviation is $\sigma = \sqrt{n \cdot p \cdot \left(1 - p\right)} = \sqrt{2 \cdot \frac{4}{17} \cdot \frac{13}{17}} = \sqrt{\frac{104}{289}} \approx 0.5999$
The special formulas $\mu = n \cdot p$ and $\sigma = \sqrt{n \cdot p \cdot \left(1 - p\right)}$ only work for binomial random variables, where $n$ is the number of independent trials with outcomes "success" or "failure", and $p$ is the probability of "success" on each trial. The random variable $X$ itself counts the number of successes in $n$ trials.