Should a Gaussian Curve Always Be Drawn Symmetrically? This question is more about the appearance of a curve that comes up while I am analysing my data rather than an actual physics question, but I think it's relevant.
I have analysed some data from an EDM experiment and am supposed to be investigating whether or not the data follows a Gaussian distribution. I have used $\chi ^{2}$ and the excel solver add-in to minimise this value and fit the resulting function over my data. 
However, when I plot my fit the curve (shown here in oragne) it isn't symmetric. It's pretty close, but not quite. Is this a problem? And if you wouldn't mind, why or why not?

 A: Here is how I interpret what happened:
You used Excel to compute the coefficients of the Gaussian that best describe the data: mean $\mu$, standard deviation $\sigma$, and magnitude $A$ for a curve
$$Y=Ae^{-(x-\mu)^2/2\sigma^2}$$
Then you evaluated that function at a number of X values. Since the X values are not symmetrical about the calculated mean, you will not see the same value at corresponding points on either side of the origin.
If you know for sure that you want the Gaussian fit in Excel to be symmetrical about X=0, it would suffice to allow Solver to only compute $\sigma$ and $A$, and set $\mu=0$.
It is quite unlikely that the fitted mean of noisy data is exactly zero: usually it is more important to test whether it could be zero: there are various statistical tests available to determine whether a particular set of observations is consistent with a particular null hypothesis (in this case, null hypothesis might be "data comes from a distribution with mean=0", and it doesn't look like your data would be sufficient to dispel that hypothesis).
Following comments on Wolphram jonny's answer, you are asking whether you can conclude that the data is Gaussian distributed. The answer is, "No you can't". It is very hard (some say, impossible) to prove something is true. You can only hope to show you can't prove it's false.
In your example, the null hypothesis would be "the data follows a Gaussian distribution". Your test would be to do a $\chi^2$ test on the fit, and see whether the value of $\chi^2$ is sufficiently small that you can't rule out that it's a Gaussian. For this you look at the p value - if it's less than some cutoff (usually 0.05) you can say (with a tip of the hat to @Henry who proposed more accurate wording):

"If this curve is in fact Gaussian, and I apply this method, I will wrongly claim it is not Gaussian (reject that hypothesis) in less than 5% (whatever p is used) of the time when I should not reject it."

The reason for the convoluted phrasing is, that random sampling from a Gaussian distribution will lead to a distribution that "looks non Gaussian" about 1 in 20 times when you use this cutoff - in other words, the p value actually says "you will get this result about p% of the time when the distribution is Gaussian".
It can be a bit confusing at first. Bottom line - your fit looks fine, don't worry about the asymmetry, continue with your chi squared test.
For my own entertainment I did the above fitting (with some made up data), with the following results:

This is the "normal" view. You can see I have entered X and Y values, and I created a "fit" column which depends on three cells (which I named mu, sigma and A); finally I created an error metric {=SUMSQ(B2:B12-C2:C12)} - note the curly brackets which you get from entering as a "array formula" (ctrl-shift-enter on PC or cmd-shift-enter on Mac). This allows you to compute the entire thing in one cell without creating a separate column with the error values. I then selected the error cell and ran the solver, minimizing cell F6 while changing cells F2:F4:

A closer look at the formulas (use ctrl-backtick to expand formulas in Excel - but note that it does not show the {} of the array formula... one of many bugs, I'm sure):

You can see here that there is a built in function =CHITEST to test the goodness of fit between the data and the Gaussian fit - and it gives a p value that is well above 0.05 so you would not be able to say this data is not normally distributed.
A: A gaussian fit is symmetrical by definition, because it is a gaussian. Your orange fit doesnt look like a gaussian, it is not even smooth. I do no think excel had a gaussian fit function (but I dont use excell so cannot tell for sure. You can use other software such as matlab, or likely free ones on the web. Or, just use that data to calculate the parameters of the best fitting gaussian and draw it on excel.
Update: I cannot tell what your professor wants, but I the fields that I have worked, when you fit a gaussian, you fit it with a continuous one (the only "real" gaussian). The data can not be symmetric, but the Gaussian will. The actual source of the data can be symmetrical, but just by chance and error your data might not be. now after you fit it with a gaussian, there are tests to tell you if the fit is good and you can blabe the asymmetry to chance, or if your data is not really well described by a Gaussian.
But I guess that will be pretty advanced for what you professor wants. I would say (just by eye) that your data  firts a Gaussian acceptably well
A: Within the specification I can glean from the question - here is what I would do.
(i) Find the best fit Gaussian - which I am assuming is what you have done.
(ii) Your best fit should return a chi-squared value
You should compare the chi-squared value with critical values of the chi-squared distribution for the appropriate number of degrees of freedom of the fit. Here I would assume you have 3 model parameters for your Gaussian (height/normalisation, width/sigma, and mean/centre), 14 data points and therefore $14-3= 11$ degrees of freedom.
e.g. For 11 d.o.f. you can reject the Gaussian hypothesis with 99% confidence if the chi-squared value exceeds 24.725.
Tables of critical values at:  http://www.medcalc.org/manual/chi-square-table.php
(iii)
Examine whether the residuals to the fit depend on $x$. If there is a trend in the residuals then although you might get an acceptable chi-squared, the trend is telling you that there is some asymmetry that is not well-fitted by a Gaussian.
Looking at your data, this doesn't seem to be the case. 
