If we assume a binomial distribution with p=35/500=0.07, q=1-p=0.93, n=500, and a critical value of 1.96 (for 95% confidence), then the margin of error is 1.96√(pq/n)=1.96√0.07×0.93/500=0.022.
The variance for a binomial distribution is npq and the standard deviation is √(npq)=√(500×0.07×0.93)=5.705 approximately.
The normal and binary distributions are approximately the same if np≥5 (µ=np=35). σ=√(np(1-p))=5.705, so B(500,0.07)∽N(35,5.705). The critical value 1.96 applied to normal distributions is then valid for the binomial distribution.