Trying to understand the math behind Avi Wigderson's simple example In his celebrated talk on randomness and pseudorandomness
https://youtu.be/Jz1UoAWD80Q?t=366
legendary mathematician Avi Wigderson makes the powerful statement that sampling is perhaps the most important use of randomness.
He skips over this rather elementary slide quickly but I am trying to understand the numbers behind it, using weak law of large numbers or Chebyhsev's inequality but I am not sure if I am getting it.
He suggests 2000 (i.i.d) samples would predict the result of an election in a 280 million population with +/-2 percent with 99% probability, he relates this to the weak law of large numbers but I can't see a concrete connection.
1/Sqrt(2000) is about 2% - which is about another thing that I never understand: why does error in direct sampling go as 1/Sqrt(N)? My feeling is error in the standard variation should go as 1/Sqrt(N) but error in the mean should scale as 1/N.
Apologies in advance, if these are too simple questions!

In choosing 2000 at random from 280,000,000 we can ignore the difference between sampling without and sampling with replacement. So, we are essentially interested in the binomial proportion $p$ of voters for Candidate A in an election contest between two candidates.
A Wald confidence interval for is based on the number $x$ in favor of A out of $n$ potential voters interviewed. We estimate $p$ by $\hat p = x/n.$ The standard
deviation of $\hat p$ is $\sqrt{p(1-p)/n}$ is called the standard error of $\hat p.$
Note that $0 \le p \le 1.$ The standard error is largest when $p = 1/2.$
The standard error itself can be estimated by $\sqrt{\hat p(1-\hat p)/n}.$

Assuming that $Z = \frac{p - \hat p}{\sqrt{\hat p(1-\hat p)/n}}$ is approximately distributed as standard normal, one has $P(-1.96 < Z < 1.96) = 0.95.$
Accordingly, one can say that an approximate 95% confidence interval for $p$
is of the form $\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}}.$ This is the Wald confidence interval and it works reasonably well as long as $n > 500$ or so.
The corresponding 99% Wald CI is of the form $\hat p \pm 2.576\sqrt{\frac{\hat p(1-\hat p)}{n}}.$ If we interview about $n = 2000$ randomly selected
potential voters, then
$$2.576\sqrt{\frac{\hat p(1-\hat p)}{n}}\le 2.576\sqrt{\frac{(1/2)(1/2)}{2000}}
= 2.576/\sqrt{8000} \approx 0.029,$$
and smaller if $p \ne 1/2;$ still smaller for a 95% CI.$
