I consider a binomial distribution with $N$ draws and the typical probabilities $p$ and $q=1-p$ and expectation value $\mu=Np$.
I now need to compute the expectation value of the function \begin{align} f(n)=\begin{cases} a (n-\mu)^3 & n<\mu\\ b (n-\mu)^3 & n>\mu \end{cases}\,. \end{align} If $a=b$, I would just compute the skewness, which is well-known for the binomial distribution, but I'm essentially interested in a skewness if the underlying function has small jump in its derivative.
Of course, I can write formally that the solution to my problem is \begin{align} \langle f\rangle=a \sum_{n<\mu} \binom{N}{n}p^n q^{N-n} (n-\mu)^3+b\sum_{n>\mu} \binom{N}{n}p^n q^{N-n} (n-\mu)^3\,, \end{align} but this is not super explicit. In fact, I would mostly like to know the leading order term and how it scales with $N$. From the definition of the skewness, I know that for $a=b$, I simply have \begin{align} \langle f\rangle=a Npq(q-p)\,. \end{align}
My question: What is $\langle f\rangle$ for $b\neq a$ up to order linear in $N$ (lower orders can be ignored)?
I tried to approach this problem by doing the equivalent calculation for a normal distribution. For this, I just assume $a=b$ for the binomial, so that for $n>\mu$, I just integrate against a Gaussian. This yields \begin{align} \langle f\rangle_{\mathrm{correction}}=(b-a)\int^\infty_{\mu}\frac{1}{\sigma \sqrt{2\pi}} e^{\frac{1}{2}(\frac{n-\mu}{\sigma})^2}(n-\mu)^3dn=(b-a)\sqrt{\frac{2}{\pi}}\sigma^3=(b-a)\sqrt{\frac{2}{\pi}}\sqrt{Npq}^3\,. \end{align} This is clearly a part of the correction (and even scales dominantly compared to the skewness), but I couldn't figure out if there is an additional correction of order $N$ due to the difference between binomial and Gaussian distribution.