In a frequency table with classes of equal width,$N_1$ is the total frequency and $N_2$ is the sum of squares of the frequencies. Assuming the distribution to be approximately normal with mean $0$, I'm asked to suggest a quick estimate of the standard deviation $\sigma$ in terms of $N_1$ and $N_2$.
My approach: I've first deduced the formula involving the sum of the square of the density of an $N(\mu,\sigma)$ variate (it doesn't depend on $\mu$) :
$\displaystyle\int_{-\infty}^{\infty}[f(x)]^2\,\mathrm{dx}=\dfrac{1}{2\sigma\sqrt{\pi}}\tag*{}$
Now, $\displaystyle\sum_{i}\dfrac{f(x)}{N_1}=1$. Then $\displaystyle\sum_{i}\left(\dfrac{f(x)}{N_1}\right)^2=\dfrac{N_2}{N_1^2}.$
I think $\displaystyle\sum_{i}\left(\dfrac{f(x)}{N_1}\right)^2\cdot h$ resembles to $\displaystyle\int_{-\infty}^{\infty}[f(x)]^2\,\mathrm{dx}=\dfrac{1}{2\sigma\sqrt{\pi}}$ where $h=$ class width.
So, $h\cdot\dfrac{N_2}{N_1^2}=\dfrac{1}{2\sigma\sqrt{\pi}}\implies\sigma=\dfrac{2h\sqrt{\pi}N_2}{N_1^2}.$
I'm confused about the answer, so please help me and suggest edits if its wrong.