Show that $E(X)=np$ when $X$ is a binomial random variable.

[Hint: First express $E(X)$ as a sum with lower limit $x=1$. Then factor out $np$, let $y=x-1$ so that the sum is from $y=0$ to $n-1$, and show that the sum equals $1$.]
 

The variance is $V(X)=np(1-p)=npq$, and the Standard Deviation (SD) of $X$ is $\sigma_X=\sqrt{npq}$ (where $q=1-p$).

The proof can be seen in Question 32. (Where the variance is denoted by $D(X)$.)