Binomial distribution central moment calculation 
If for a binomial distribution the mean is $4$ and variance is $3$, find th $3^{\text{rd}}$ central moment. 

I understand that the first and second central moments are mean and variance respectively, but what is the $3^{\text{rd}}$? A Google search only reveals formulas with $n$ in them, which is information that is, as per the question, unavailable. The answer to the  question is $3/2$, but I can't figure out where it comes from.
 A: Recall that if $X\sim\mathrm{Bin}(n,p)$, then $\mathbb E[X] = np$ and $\mathrm{Var}(X)=np(1-p)$. Given $\mathbb E[X] = 4$ and $\mathrm{Var}(X) = 3$, we have $np = 4$ and $np(1-p)=3$. Hence $n=16$, $p=\frac14$. So the distribution of $X$ is given by
$$\mathbb P(X=k) = \binom {16}k \left(\frac14\right)^k\left(\frac34\right)^{16-k}, k=0,1,\ldots,16.$$
The second moment of $X$ is
$$\mathbb E[X^2] = \mathrm{Var}(X) + \mathbb E[X]^2 = 3 + 4^2 = 19.$$
The generating function of $X$ is
$$\begin{align*}
\mathbb E[s^X] &= \sum_{k=0}^\infty\mathbb P(X=k)s^k.
\end{align*}$$
Since $X$ is the sum of $16$ i.i.d. $\mathrm{Ber}(p)$ random variables, each with generating function $1-p+ps$, we have
$$\mathbb E[s^X] := P(s) = (1-p+ps)^n = \left(\frac34 + \frac14s\right)^{16}.$$
The third central moment of $X$ is
$$\begin{align*}
\mathbb E[(X-\mathbb E[X])^3] &= \mathbb E[(X-4)^3]\\ 
&= \mathbb E[X^3 - 12X^2 + 48X - 64]\\
&= \mathbb E[X(X-1)(X-2) - 9X^2 + 46X - 64]\\
&= \mathbb E[X(X-1)(X-2)] - 9\mathbb E[X^2] + 46\mathbb E[X] - 64.
\end{align*}$$
Now,
$$\begin{align*}\mathbb E[X(X-1)(X-2)] &= \lim_{s\uparrow1}P^{(3)}(s)\\ 
&= \lim_{s\uparrow1}\left(\frac14\right)^3 16\cdot15\cdot14\left(\frac34+\frac14s\right)^{13}\\
&= \frac{105}2. \end{align*}$$
Hence
$$\begin{align*}
\mathbb E[(X-\mathbb E[X])^3] &= \mathbb E[X(X-1)(X-2)] - 9\mathbb E[X^2] + 46\mathbb E[X] - 64\\
&= \frac{105}2 - 9\cdot19 + 46\cdot 4 - 64\\
&= \frac32.
\end{align*}$$
A: By definition, the third central moment of a random variable $X$ is $E[(X-\mu)^3]$ where $\mu = E[X]$.  
The Binomial($n,p$) random variable $X$ can be written as the sum of $n$ independent Bernoulli($p$) random variables $Y_i$, but we want to subtract the mean, so write 
$ X - np = \sum_{j=1}^n Z_j$ where $$Z_j = Y_j - p = \cases{1-p & with probability $p$\cr -p & with probability $1-p$\cr}$$
are independent.  Now
$$ E[(X - np)^3] = \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n E[Z_i Z_j Z_k]$$
There are $n$ terms where $i=j=k$, and $E[Z_i^3] = p (1-p)^3 - p^3 (1-p)$.
The other terms are either of the form $E[Z_i Z_j^2]$ with $i \ne j$ or $E[Z_i Z_j Z_k]$ with $i,j,k$ all different, and those are all $0$ since 
the $Z$'s are independent and have mean $0$.  So we conclude that the 
third central moment is 
$$ E[(X -np)^3] = n (p (1-p)^3 - p^3 (1-p))$$
which can be simplified to $n p (1-p)(1-2p)$.
A: When it comes to central moments, it is often much easier to use the cumulant-generating function $K(t)$, which is the natural log of the moment-generating function. In this case the function is
$$ K(t) = \ln (1 - p + pe^{t})^{n} = n\ln (1 - p + pe^{t}) $$
The mean, the variance, and the third central moment, for example, are obtained by taking respectively the first, the second, and the third derivatives of the cumulant function with respect to $t$ and evaluating them at 0.
A: In this question 
Np = 4 mean (1)
Npq = 3 variance (2)
Solving from (1) and (2) 
$P = 1/4$ 
$Q = 3/4$ 
$N = 16$ 
This bionomail 1 moments = 0 
2 moments = npq ie 3 
3 moment = npq(q-p) = 3 X 0.5 = 3/2 
4 moment = 3 ( npq)^2 + npq (1-6pq ) = 3(3) ^2 + 3 ( 1- 6 ( 0.1875) = 
= 26.625
