Zwitterion (amino acid) equilibrium and the equilibrium constant
If you assume that the groups react independently, you have two equilibria with equilibrium constants $K_a(\ce{-NH3+})$ and $K_a(\ce{-COOH})$:

$$\ce{^+(H3N)-R &lt;=&gt; (H2N)-R + H+}$$

and

$$\ce{R-COOH &lt;=&gt; R-COO- + H+}$$

You are interested in the overall reaction

$$\ce{^+(H3N)-CHR-COO- &lt;=&gt; (H2N)-CHR-COOH}.$$

You can take a two step route to achieve this:

$$\ce{^+(H3N)-CHR-COO- &lt;=&gt; (H2N)-CHR-COO- &lt;=&gt; (H2N)-CHR-COOH}$$

The equilibrium constant for the first step is $K_a(\ce{-NH3+})$, and for the second step is $1/K_a(\ce{-COOH})$. Combined, you get a result of

$$ K_\text{combined} = \frac{K_a(\ce{-NH3+})}{K_a(\ce{-COOH})}$$

Because the $K_a$ values of ammonium and carboxylic acid are different by about four orders of magnitude, the uncharged species occurs about ten thousand times less often than the zwitter ionic form in aqueous solution. In general, comparing p$K_a$ values of two groups competing for a proton tells you which group will usually have the proton (the one with the larger p$K_a$) and how unlikely a proton transfer to the other group would be ($10^{\Delta pK_a}$).


  What are the faults (calculation errors, false/inappropriate assumptions, etc) in my working, if any?


I don't understand line 2 and 3 of your argument, and how you get from line 5 to 6. The result you get differs from mine in a square root.


  I thought zwitterions only consist a small portion of the mixture


This depends on the solvent. In aqueous solution, zwitterions are the common species, whereas in organic solvent, the uncharged species is the common one.