We can take many approaches to arrive at the solution. A conceptually and mathematically elegant one begins by characterizing the skewness and kurtosis as properties of the empirical distribution. This is the distribution of the random variable $X$ defined by putting $n_1$ tickets labeled with the number $x_1,$ $n_2$ tickets labeled with $x_2,$ and so on, into a box and withdrawing one of those tickets at random. This latter phrase means each ticket has the same chance of being withdrawn. Consequently, the chance $p_i$ of observing $x_i$ (for any $i$) must be the number of tickets with $x_i$ on them (equal to $n_i,$ provided all the $x_i$ are distinct) divided by all the tickets. Thus, when there are $d$ distinct values on the tickets,
$$p_i = \frac{n_i}{n_1+n_2+\cdots+n_d}.$$
The rest is a matter of applying the definitions.
The mean (aka "expectation") is the sum of the values times their probabilities, $$\mu_1 = E[X] = \sum_{i=1}^d p_i x_i.$$
For any $k=1,2,3,\ldots,$ the $k^\text{th}$ central moment is the expectation of $(X-\mu_1)^k:$ $$\mu_k = E[(X-\mu_1)^k] = \sum_{i=1}^d p_i (x_i - \mu_1)^k.$$
For any such $k,$ the standardized central moment $\beta_k$ is the expectation of $Z^k$ where $Z = (X-\mu_1)/\sqrt{\mu_2}.$ With a little algebra this simplifies to $$\beta_k = \frac{\mu_k}{\mu_2^{k/2}}.$$
The skewness is $\beta_3$ and the kurtosis is $\beta_4.$ (Sometimes "kurtosis" refers to the "excess kurtosis," which is $\beta_4 - 3.$)
Example
Here is a simplified version of the data in the question, where the counts have been reduced so the arithmetic details are less distracting.
$$\begin{array}
&i & x_i & n_i \\
\hline
1 & 100 & 1\\
2 & 101 & 9\\
3 & 102 & 6\\
4 & 103 & 3\\
5 & 104 & 1
\end{array}$$
These are the calculations:
- $n_1 + \cdots + n_d = 1+9+6+3+1=20.$
- $p_1=1/20, p_2=9/20, p_3=6/20, p_4=3/20, p_5=1/20.$
- $\mu_1 = \frac{1}{20}(100) + \frac{9}{20}(101) + \frac{6}{20}(102) + \frac{3}{20}(103) + \frac{1}{20}(104-101.7)^2 = 101.7.$
- $\mu_2 = \frac{1}{20}(100-101.7)^2 + \cdots + \frac{1}{20}(104-101.7)^2 = 0.91.$
- $\mu_3 = \frac{1}{20}(100-101.7)^3 + \cdots + \frac{1}{20}(104-101.7)^3 = 0.546.$
- $\mu_4 = \frac{1}{20}(100-101.7)^4 + \cdots + \frac{1}{20}(104-101.7)^4 = 2.3557.$
- $\beta_3 = 0.546 / (0.91^{3/2}) \approx 0.628971.$
- $\beta_4 = 2.3557 / (0.91^{4/2}) \approx 2.8447.$
In many computing platforms it is convenient to compute the residuals $x_i-\mu_1$ once and for all because these terms recur in all the formulas for the $\mu_k.$ This leads to easy spreadsheet formulas and to straightforward code, such as this R example of the preceding calculations. Their connections to the foregoing formulas should be obvious.
x <- 100:104
n <- c(1,9,6,3,1)
p <- n / sum(n)
mu.1 <- sum(p*x)
r <- x - mu.1
mu.2 <- sum(p*r^2)
mu.3 <- sum(p*r^3)
mu.4 <- sum(p*r^4)
beta.3 <- mu.3 / mu.2^(3/2)
beta.4 <- mu.4 / mu.2^(4/2)
