I was asked the following:
Estimate the values $\mu$ and $\sigma$, from the following diagram:
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 
I do get that the $\mu = 64$. I didn't know how to calculate the $\sigma$-value. The answer was $\sigma=1$.
Here is my question: Why? I have no idea how to get to that conclusion. Can you please help me?
Thanks a lot in advance!
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{\fermi\pars{x,\sigma} \equiv 5\exp\pars{-\,{\pars{x - 64}^{2} \over 2\sigma^{2}}}}$ $$ \mbox{the minimum of}\ \,{\cal F}\pars{\sigma} \equiv \bracks{\fermi\pars{65,\sigma} - 3}^{2} + \fermi^{2}\pars{68,\sigma} $$ is found with $\ds{\color{#66f}{\large\sigma} \approx \color{#66f}{\large 0.989344}}$
With a 'drastic' $\ds{\fermi\pars{65,s} = 3}$, we'll get $\dsc{0.989347}$.