A Survey of Confidence

The margin of error of a confidence interval for a population proportion is given by

$m = z^* \sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}},$

where $z^*$ is the critical value (depending on the confidence level), $\hat{p}$ is the sample proportion, and $n$ is the sample size. Since we do not know an approximate value for the proportion of Brilliant users that like physics, we take $\hat{p} = 0.5,$ since that maximizes the margin of error. Thus, we have

$\begin{aligned} z^* \sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}} &\leq 0.02 \\ 1.96 \sqrt{\dfrac{0.5^2}{n}} &\leq 0.02 \\ \dfrac{1.96(0.5)}{\sqrt{n}} & \leq 0.02 \\ n &\geq \left ( \dfrac{1.96(0.5)}{0.02} \right )^2 \\ n &\geq 2401. \end{aligned}$

This means Calvin needs to survey at least $\boxed{2401}$ Brilliant users in order to have a margin of error less than 2%.