No, those equations come directly from the mean and variance formulae in terms of expected value, considering the colletedcollected data as a population.
$$\mu = \mathbb{E}\big[X\big]$$
$$\sigma^2 = \mathbb{E}\big[\big(X-\mu\big)^2\big]$$
Since you have a finite number of observations, the distribution is discrete,$^{\dagger}$ and the expected value is a sum.
$$\mu = \mathbb{E}\big[X\big] = \sum_{i=1}^N p(x_i)x_i = \sum_{i=1}^N \dfrac{1}{N}x_i = \dfrac{1}{N}\sum_{i=1}^Nx_i$$
$$\sigma^2 = \mathbb{E}\big[\big(X-\mu\big)^2\big] = \sum_{i=1}^N p(x_i)(x_i - \mu)^2 = \sum_{i=1}^N \dfrac{1}{N}(x_i - \mu)^2 = \dfrac{1}{N}\sum_{i=1}^N (x_i - \mu)^2$$
(To get from $p(x_i)$ to $\dfrac{1}{N}$, note that each individual $x_i$ has probability $1/N$.)
This is why the $\dfrac{1}{N}\sum_{i=1}^N (x_i - \mu)^2$ gets called the "population" variance. It literally is the population variance if you consider the observed data to be the population.
$^{\dagger}$ This is a sufficient, but not necessary, condition for a discrete distribution. A Poisson distribution is an example of a discrete distribution with infinitely many values.
