For a natural number $n \in \mathbb{N}$, let $P_{n,p}$ be the probability mass function of the binomial distribution, that is, for $i \in [0,n]$, $P_{n,p}(i) = \binom{n}{i} p^i (1-p)^{n-i}$.
We are interested in the scenario where $p$ is fixed from the beginning and $n$ grows.
The binomial distribution concentrates around the expectation value $p\cdot n$ and lot of concentration inequalities for the binomial distribution are known: For example, we know that for any fixed $C>0$ and $i \in [pn - C\sqrt{n}, pn + C\sqrt{n}]$ we have $P_{n,p}(i) = \Omega(\frac{1}{\sqrt{n}})$. (This can be seen using Stirlings approximations formula.)
First question: Let $\mathcal{I} =[pn - C\sqrt{n}, pn + C\sqrt{n}]$. Is it true that for any fixed $\delta >0$, we can find $C>0$ such that $\sum_{i \in \mathcal{I}} P_{n,p}(i) \geq 1 - \delta$ for all $n$ sufficiently large?
Second question: Using $P_{n,p}(i) = \Omega(\frac{1}{\sqrt{n}})$ for $i \in \mathcal{I}$, it follows that $\sum_{i \in \mathcal{I}} P_{n,p}(i) = \Omega(1)$. Is $\mathcal{I}$ the smallest such interval? Or can we find some $f$ which grows slower than $\mathcal{O}(\sqrt{n})$ such that $\sum_{i \in \mathcal{J}} P_{n,p}(i) = \Omega(1)$ when $\mathcal{J} = [pn - f(n), pn + f(n)]$?
The first answer of this math overflow question gives a bound for terms like $\sum_{i \leq k} P_{n,p}(i)$ which might help to solve the questions, but I am unfortunately unable to solve it for $k$.