$m\sim$Bin(3000, 1/6). Find
$$P\left(\left|\frac{m}{3000}-\frac{1}{6}\right|\right)\le 0,01.$$
So far, I know this formula, but I have been told,
$$x_0=\epsilon \cdot \sqrt{\frac{n}{pq}} \Rightarrow x_0 = 0,01 \cdot \sqrt{\frac{3000}{\frac{1}{6}\cdot \frac{5}{6}}}=60,$$ but I don't know what $x_0$ even stands for, and why it is calculated the way it is?
Then from Gaussian table we get that $$\int_0^{60}f(x)dx=1$$. Here of course I may have made an error, but on Gaussian table every area that is beyond 3 is equal to one, so I thought at least. But then $$P\left(\left|\frac{m}{3000}-\frac{1}{6}\right|\le 0,01 \right) =2 \cdot 1=2.$$. But of course such probaibility can't be true. So, where have I made a mistake?