The sums of squares from standard deviation

Question

I am trying to figure out how to get a sum of squares from a standard deviation. The standard deviations I have are $11.04$, $9.91$ and $9.43$. How do I calculate the sum of squares associated with these standard deviation values? Thanks!

Answer 1

It holds that $$ \sigma^2 = \frac{1}{n}\sum_{i=1}^n (x_i-\bar{x})^2 = \Big[\frac{1}{n}\sum_{i=1}^n x_i^2\Big] - (\bar{x})^2 $$

hence we have $$ \sum_{i=1}^n x_i^2\ = n[\sigma^2+(\bar{x})^2]. $$

Note that $\bar{x}$ is arithmetic mean and $n$ is number of observation.