Let p denote the percentage of all 17-year-old students in the U.S who knew the right answer. An estimate of p is given by 617/950=.65 or 65% of the SRS sample.

The estimate of the variance of p is given by s^2(p)=(estimated p)*(1-estimated p)/n. So, the estimated standard error of the estimate of p is SE(p)=.015. Due to the large sample (n*estimated p>5) and n*(1-estimated p)>5) from the Central Limit Theorem we get that p is approximately normally distributed with mean=.65 and variance=.00024. Hence, a 95% confidence interval is given by (.65-1.96*.015, .65+1.96*.015)=(.619, .679).

If they wanted to be 95% sure of getting the national percentage p right to within 2 percentage points, they would have to calculate the following:

P( | p-estimated p | <= .02)=.95 or standardizing

P([ | p-estimated p | / SE(p) ] <= a )=.95

where a=.02/SE(p). But a=1.96, which implies that (.02)/[(estimated p)*(1-estimated p)]/n=1.96. Solving the last equation for n and using estimated p=.65 we get that n=2185.