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 $D(X)$.)
1
Suppose $X\sim\mathrm{Bin}(n,p)$, Then
\[
\begin{split}
E(X)&=\sum_{x=0}^{n}xp(x)=\sum_{x=0}^{n}x\cdot b(x;n,p)\\
&=\sum_{x=0}^{n}x\cdot\binom{n}{x}p^x(1-p)^{n-x}\\
&=\sum_{x=1}^{n}x\cdot\binom{n}{x}p^x(1-p)^{n-x}\\
&=np\sum_{x=1}^{n}\frac{1}{n}x\cdot\binom{n}{x}p^{x-1}(1-p)^{n-x}\\
&=np\sum_{x=1}^{n}\frac{1}{n}x\cdot\frac{n!}{x!(n-x)!}p^{x-1}(1-p)^{n-x}\\
&=np\sum_{x=1}^{n}\frac{(n-1)!}{(x-1)!\bigl((n-1)-(x-1)\bigr)!}p^{x-1}(1-p)^{(n-1)-(x-1)}\\
\end{split}
\]
Let $y=x-1$, then we get
\[
\begin{split}
E(X)&=np\sum_{y=0}^{n-1}\frac{(n-1)!}{y!\bigl((n-1)-y\bigr)!}p^{y}(1-p)^{(n-1)-y}\\
&=np\sum_{y=0}^{n-1}\binom{n-1}{y}p^{y}(1-p)^{(n-1)-y}\\
&=np\cdot\bigl(p+(1-p)\bigr)^{n-1}\\
&=np\cdot 1\\
&=np
\end{split}
\]
Therefore, the expected value of the binomial random variable is equal to $np$.