If Σ_i=1^n (x_i- a )= n and Σ_i=1^n (x_i- a )^2= na, ( n, a >1) then the standard deviation of n observations x_1, x_2, …, x_n is:

If $\displaystyle \sum_{i=1}^{\mathrm{n}}\left(x_i-\mathrm{a}\right)=\mathrm{n}$ and $\displaystyle \sum_{i=1}^{\mathrm{n}}\left(x_i-\mathrm{a}\right)^2=\mathrm{na}$, $\displaystyle (\mathrm{n}, \mathrm{a}>1)$ then the standard deviation of n observations $\displaystyle x_1, x_2, \ldots, x_{\mathrm{n}}$ is :