How to find confidence interval of 0.95 in this problem? For sample $x_1,\cdots,x_{100}$, following holds.
$\sum_{k=1}^{100}x_k=400$ and $\sum_{k=1}^{100}x^2_k=2500$.
Find the confidence interval of 0.95 for the population mean $m$.

I've calculated the sample mean, $\bar{x}=4$, and I know that $\bar{x}-1.96\sigma n^{-1/2}\leq m \leq \bar{x}+1.96\sigma n^{-1/2}$. The values are $n=100$, $\bar{x}=4$ but I don't know how to get $\sigma$. I think I have to use the sum of $x^2_k$ to get $\sigma$ but can't figure it out.
You will never get $\sigma$ because $\sigma$ denotes the population standard deviation which is unknown. You will get instead $s$ which denotes the sample standard deviation and is equal to the square root of the sample variance, i.e. $s=\sqrt{s^2}$, where $$s^2=\frac{1}{n-1}\left(\sum_{k=1}^{n} x_k^2-\frac{1}{n}\left(\sum_{k=1}^{n}x_k\right)^2\right)$$
Hint: Use the formula: sigma = ((100*2500 - 400^2)/(100*(100 - 1)))^(1/2) = 3.015