standard model - Who chooses the representation of SM particles?

I refer to this wikipedia table where the SM particles and their transformation behaviour are listed. When considering the field $W$ e.g., the 'representation' is listed as $(\mathbf {1} ,\mathbf {3} ,0)$.

So far my understanding of that is, that $W$ does not transform under $SU(3)$, transforms as a triplet under $SU(2)$ (means it transforms in the adjoint) and does not transform under $U(1)$.

The transformation in some representation of the gauge bosons make (at least a bit of) sense to me, since it has to keep the lagrangian invariant as the fermions transform.

However, the transformation of the fermions seem to have an arbitrary transformation behaviour to me. Is there any reason why we have a fermion field that transforms as $({\bar {\mathbf {3} }},\mathbf {1} ,\textstyle {\frac {2}{3}})$ (left handed downtype anti fermion) but no $({\bar {\mathbf {7} }},\mathbf {3} ,-\textstyle {\frac {11}{3}})$ fermion field?

If the answer is "because it describes what we see": Why does the number of fields and their transformation behaviour then not count as free parameters of the model?

Answer

The number of fields and their representations are parameters you can freely choose when building a quantum field theory. The only constraints the fermionic representations under the gauge symmetries must obey is that they must be free of a total anomaly for the gauge symmetry, but this is actually not a very strong constraint.

The Standard Model is not "special", nothing about its field content is particularly noteworthy or unique among all possible quantum field theories as far as we know. It is not derived from theoretical considerations of consistency or beauty - the Standard Model with its specific field content simply arises from the best fit to observational data.

If you want the field content to be less arbitrary, you need to add stronger constraints than "being a quantum field theory". For instance, the strongest constraint known is "being a theory of supergravity in high dimensions". We cannot have particles of spin greater than 2 in our theory due to no go theorems sometimes called the "Weinberg-Witten theorem", and supergravity in more than 11 dimensions inevitably would contain those. In fact, there is exactly one theory of supergravity in eleven dimensions - thought to be the low-energy description of M-theory - and five of them in ten dimensions, corresponding to the low-energy descriptions of the five different string theories. Unfortunately, this remarkable uniqueness does not extend to our four dimensions, where the supersymmetric QFTs are still somewhat more constrained, but not so strongly.