Standard deviation of gaussian noise in FHEW scheme
Apologies for seeing this question just now. For your first question, well, there seems to be a typo: I should have written $\|s\|^2 \leq n/2$, and then we have $\sqrt{(\|s\|^2 +1) /12 }  = \sqrt{251/12} \approx 4.57$.

Now, we take 2 ciphertext with LWE error of std-deviation $\beta$, and sum them. Assuming independence the std-dev of the sum of the errors is $\sqrt 2 \beta$. For correctness, we need this sum of errors to be less than $q/8$ (Here, to save every last bit, we relax the naive condition stated in Lemma 7, and directly bound the sum $q/8 + |e_0 + e_1|$ by $q/4$ rather than bounding $e_0$ and $e_1$).

answer to comment on decreasing error via swapping KeySwitch and $\textsf{HomNAND}$

Just for the sake of this explanation, let me lie about the actual noise propagation, and assume that each operation HN and KS just add a constant to the quantity of noise, denoted $h$ and $k$.

The ``easy to describe scheme'' does $\textsf{HomNAND}(KS(X_1), KS(X_2))$, which leads to an error of the form $(x_1 + k) + (x_2 + k) + h = x_1 + x_2 + h + 2k$.

The ``better but less easy to describe scheme'' does $KS(\textsf{HomNAND}(X1, X_2)) $which leads to and error $(x_1+x_2+h) + k$. That is one ks less. 

It turns out that even with the actual formula, we also gain a little bit on the final error with this trick.