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!
I think you are supposed to use that about 68% of values drawn from a normal distribution are within one standard deviation $\sigma$ away from the mean; about 95% of the values lie within two standard deviations and about 99.7% are within three standard deviations. See http://en.wikipedia.org/wiki/File:Empirical_Rule.PNG for a link to a usefull picture.
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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 $\ds{\color{#66f}{\large\sigma} \approx \color{#66f}{\large 0.989344}}$
Here we can see a plot of $\ds{\,{\cal F}\pars{\sigma}}$ where $\ds{\sigma \in \bracks{0.8,1.2}}$:
With a 'drastic' $\ds{\fermi\pars{65,s} = 3}$, we'll get $\dsc{0.989347}$.
Here is my question: Why? I have no idea how to get to that conclusion.
The overwhelming majority of the surface determined by the graphic of the Gaussian bell curve is supposed to be comprised in the interval $\mu\pm3\sigma$. That's what the $\sigma$ is for. Looking at your graphic, we notice that the overwhelming part of its entire area in comprised within the interval $64\pm3$. The conclusion should be obvious.
Simple, in a Normal distribution, $\mu$ is the intersection of the symmetry axis with $x$ axis and $\sigma$ is the distance between the symmetry axis and the inflection point on both sides of the Normal, which visually appears to be 1 (length of the side of the reference squares in the figure). See this site for details on how to find the inflection points (link).
