The issue is that making a histogram with just 3 unique numbers is just plain silly.

sig = 5*10^-4;
mn = 162.7;

dat = RandomReal[NormalDistribution[mn, sig], 1000];

Table[
 (*function of normal distribution*)
 fun = PDF[NormalDistribution[u, s]][x];
 mn = Mean[dat];
 std = StandardDeviation[dat];

 {bin, cnt} = HistogramList[dat, {d std}, "PDF"];
 cent = MovingAverage[bin, 2];
 cDat = Transpose[{cent, cnt}];

 (*find fit using the points*)
 sol = fun /. FindFit[cDat, fun, {{u, mn}, {s, std}}, x];

 (*use build in function*)
 dist = PDF[EstimatedDistribution[dat, NormalDistribution[u, s]]][x];

 Show[
  h = Histogram[dat, {(d std)}, "PDF"],
  ListPlot[cDat],
  (*use the range of the histogram for your plot*)
  {min, max} = Options[h, PlotRange][[1, 2, 1]];
  Plot[{sol, dist}, {x, min, max}]
  , ImageSize -> 500]
 , {d, .1, .5, .1}]

Yes i understand, but i just generated some data with the same properties, the code will also work on your data, just replace the artificially generated data with your data.

It was more to demonstrate how you can automatically propagate plot ranges etc. the only input of my code is the bin width which is a percentage of the std.

By estimating all the ranges from the data the chance of "copy paste" / "type errors" is minimized. As you can see the mn and std i use to generate the data are not used anywhere else (although i only now see i reuse and overwrite the mn parameter).

Oh my, I made a mistake. I was making multiple histograms and I copied over the wrong data. I guess that is what hapens when you make a post late in the evening. I edited it to the right values.

I am beyond words both in how thankful and disappointed in myself I am. For some reason the PlotRange fixes everything. Thank you so much.