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Let's first test which distribution is more likely. However, we should treat FindDistribution results with a grain of salt, it may be suggestive, but there are problems with the definition of AIC, which is not industry standard, e.g., see https://community.wolfram.com/groups/-/m/t/2198153 and prior links to it.

What I did was ignore the binning with the command data[[All, 2]], and doing that doesn't really harm anything other than shift the data by about 23 to the left. This shows us that a normal distribution is likely good enough, and it would take a 3 parameter Weibull distribution to be competitive. Now this may be a three parameter Weibull distribution. In any case, treating it like a normal distribution one can

Let's first test which distribution is more likely. However, we should treat FindDistribution results with a grain of salt, it may be suggestive, but there are problems with the definition of AIC, which is not industry standard, e.g., see https://community.wolfram.com/groups/-/m/t/2198153 and prior links to it. Worse, this is the wrong routine to use for this data (mismatch), but I'm just mucking with the data to get ideas as to what might be worth testing.

What I did was ignore the binning with the command data[[All, 2]], and doing that doesn't really harm anything other than shift the data by about 23 to the left. This suggests that perhaps a normal distribution may be good enough, and it would take a 3 parameter Weibull distribution to be competitive. Now this may be a three parameter Weibull distribution. In any case, treating it like a normal distribution one can do another thing that seems like heresy:

Let's first test which distribution is more likely.

Let's first test which distribution is more likely. However, we should treat FindDistribution results with a grain of salt, it may be suggestive, but there are problems with the definition of AIC, which is not industry standard, e.g., see https://community.wolfram.com/groups/-/m/t/2198153 and prior links to it.

Which gives us From the QQ plots, we can see the problem with just duplicating samples, we get a dotted step pattern $...\cdots$. Despite this, it is quite clear that the Normal distribution (see left tail) fits slightly better than the 3 parameter Weibull, which in turn fits much better than the 2 parameter Weibull. In doing so, we are not so much concerned with getting an exact answer for each fit, but are answering the question as to which fit equation is appropriate.

Which gives us From the QQ plots, we can see the problem with just duplicating samples, we get a dotted step pattern $...\cdots$. Despite this, it is quite clear that the Normal distribution (see left tail) fits slightly better than the 3 parameter Weibull, which in turn fits much better than the 2 parameter Weibull. In doing so, we are not so much concerned with getting an exact answer for each fit, but are answering the question as to which fit equation is appropriate. To see how to get reproducible answers that are largely methods independent, follow the procedure in the chat link. https://chat.stackexchange.com/rooms/137349/room-for-jimb-and-carl