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I'm working on Franck–Hertz experiment for labs at my university. My professor suggested using normal distribution to find the two maxima of function I(U) (see figure below).

enter image description here

I did fine with finding the maxima obtaining the following parameters for normal distribution for one of them: $$\mu=54.7712 \ [52.6229, 56.9195] $$ $$\sigma=4.59029 \ [3.49087, 6.70445] $$ The intervals next to the parameter estimates are the 95% confidence intervals for the distribution parameters. So $\mu$ is my maximum point but how do I find its uncertainty? Is it $\sigma$? Or do I need to do something with the intervals? I've never used this method and I'm confused.

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  • $\begingroup$ If I follow you, you fit each each of the curves around a peak with a normal distribution. Your example shows three peaks but you imply there are two; can you clarify? Each of the curves does not look normal, they look skewed to to the right, so why use a normal (not skewed) distribution? Regardless of what distribution you use to fit a peak, that distribution has a mean, a standard deviation, and percentiles, so I do not understand why you have percentiles for the mean and for the standard deviation? Can you please provide more details? I am confused. $\endgroup$
    – John Darby
    Commented Nov 26, 2020 at 4:12
  • $\begingroup$ I've changed the figure so that it shows my results to avoid any confusion. Yes, I'm trying to fit normal distibution around these peaks to find for what x we obtain the maximum value. The number of peaks doesn't really matter, however now you can clearly see my results have two maxima fully visible. The $\mu$ and $\sigma$ parameters in the original post are for the 2nd maximum. So $\mu$ is my x value where the original curve has a maximum but I need its uncertainty and I don't know whether it's the $\sigma$ parameter or not. I'm not sure about the percentiles, that's just Matlab output. $\endgroup$
    – mcas
    Commented Nov 26, 2020 at 13:47
  • $\begingroup$ have you tried using mathworks.com/help/stats/prctile.html $\endgroup$ Commented Nov 26, 2020 at 13:50
  • $\begingroup$ I'm not sure how it would help me as I have no problem with calculating things but understanding what I'm actually calclulating. I'll try to clarify what my problem is with this example: $\endgroup$
    – mcas
    Commented Nov 26, 2020 at 13:58
  • $\begingroup$ I can fit a quadratic function into these maxima as well. So I find its parameters a, b and c and take the derivative so I can find its maximum. So the maximum is $x=\frac{-b}{2a}$. I need its uncertainty so it's $u_c(x)=\sqrt{(u(b)\cdot\frac{\partial x}{\partial b})^2+(u(a)\cdot\frac{\partial x}{\partial a})}^2$. And now, what I can't do is calculate the uncertainty from fitting with normal distrubution. $\endgroup$
    – mcas
    Commented Nov 26, 2020 at 14:05

1 Answer 1

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The short answer is that $\mu$ is the center and $\sigma$ is the width of the Gaussian curve which you are fitting. Thus, $\sigma$ has nothing to do with the estimate of the peak.

Rather, since you have provided incomplete data, MatLab has done its best to guess how a fitted Gaussian would look. It has provided you with confidence intervals, and the easy solution is to just say that with 95% confidence you have $\mu = 54.77 \pm 2.15$, depending on how many significant digits you want. If you add / subtract the uncertainty you get the endpoints of the interval MatLab gave you.

To better clear up your confusion, you should take a look at how MatLab calculates that interval and try to understand where it comes from.

Also, while the normal distribution and the Gaussian curve are related, it is not entirely accurate to say you fit a normal distribution here. The data points are not a statistical distribution but a plot of a function.

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  • $\begingroup$ So you fit the data with a normal distribution using MatLab. If I understand @Codename 47 answer, the uncertainties are the uncertainties for the curve fit. I believe $\sigma $ is the standard deviation for the fitted distribution. $\endgroup$
    – John Darby
    Commented Nov 26, 2020 at 14:53
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    $\begingroup$ @JohnDarby The data is current as a function of voltage $I(U)$, not a statistical distribution. They would like to identify the peaks of this function even though his data might not include that peak. Thus they approximate each peak as a Gaussian function, implicitly interpolating between the data points to find the maximum. What they want to know is the uncertainty involved in this interpolation. That has nothing to do with the width of the fitted Gaussian. $\endgroup$ Commented Nov 26, 2020 at 15:19

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