A survey company wants a margin of luck of 2 percentage points…

The margin of luck of a fair random poll is about $\dfrac{100}{\sqrt n}$ percentage points, for about 95 polls in 100. For a margin $m$, take $n = \left(\dfrac{100}{m}\right)^2$. A group of $n_g$ inside the poll has the margin $\dfrac{100}{\sqrt{n_g}}$.