How can I measure which function is a better fit to a set of datapoints The problem is from here. Namely, I have set of data points
data = [[90.00, 2.0], [97.40, 5.0], [104.8, 14.0], [112.2, 12.0], [119.6, 11.0], [127.0, 6.0], [134.4, 3.0], [141.8, 1.0], [149.2, 2.0], [156.6, 1.0]]

I have to fit a curve of the form $$\frac{m}{\sigma\sqrt{2\pi}}\cdot e^{-(x-\mu)^2/(2\sigma^2)}$$
to this data.
In the link, Volker Brown got the values 
$$m = 405.75796954829985, \mu = 111.86913960269014, \sigma = 11.968861052746961$$
and as I tried my version of Sage to the same to that set of data, I got
$$m = 405.7572223708457, \mu = 111.86914614035226, \sigma = 11.96881745593664$$
What kind of ways there are to measure which one is better fit to that set of data? I heard that sometimes one can use log-likelihoods to select the best model and sometimes Akaike's information criterion is suitable for that. But which one should I choose in this situation? Thusfar, I haven't studied the theory behind the AIC. 
Another thing that confuses me is Brown's comment "You need to give some hint for the initial values." Does that mean that there is no standard way to measure how good a particular curve fits to the data?
Or is that just a bug in a previous version of Sage that has been fixed in the version  5.6, 21 January 2013?
 A: You should note that AIC is just a penalized liklihood where the penalty is based on the number of parameters fit.  Since the 2 models that you are comparing both fit the same number of parameters (just found different values) they will both receive the same exact penalty.  So for your case comparing AIC values will give the same results as comparing liklihoods.
The advantage of AIC is when comparing models with different numbers of parameters (i.e. a linear model with additional covariates, interaction terms, polynomial terms, etc.).  The more complicated model will tend to have a higher liklihood, possibly due to random chance.  The AIC value looks to see if the increase of liklihood is enough to justify the added complexity.  Since your 2 models are equally complex the liklihood part is the only thing that will change.
The comment about the starting values is because many algorythms are sensitive to starting values.  You may have found a local max/min or the process may have stopped because the last improvement was small compared to previous steps.  Think about the question "Where is your home?", if you are asked this while in your own neighborhood you are likely to give your street name and possibly the exact address, if in a different country you are more likely to only give the city, or possibly even a nearby city that is more likely to be recognized.  Similarly the closer your starting values are to the "Truth" the stricter the criteria for stopping.
Have you tried actually plotting your data and the 2 fitted equations?  Given that the 2 sets of parameters don't differ until after the 7th significant digit, I doubt that you will see any practical difference.  That level of difference is easily explained by differences in starting values rather than bugs or changes in the software  (rarely do we ever get the exactly correct answer, just reasonably good approximations).
