Question:
How many times do we have to throw a normal coin so that, we are at least $99\%$ sure, that the percentage of heads will be between $49.5\%$ and $50.5\%$?
Solution:
$$\begin{align} 99\% \ \text{CI} &= p \pm 2.575 \cdot \sqrt{\frac{p(1-p)}{n}} \\ (0.495, 0.505) &= 0.5 \pm 2.575 \cdot \sqrt{\frac{0.25}{n}} \\ 0.005 &= 2.575 \cdot \frac{0.5}{\sqrt{n}} \\ n &= \left(2.575 \cdot \frac{0.5}{0.005}\right)^2 \\ &= 66306.25 \end{align}$$
then $66307$ tosses would be required.
Could you please explain the solution in a simple way? Or if it's possible to give an alternative solution?