This is a binary situation. We use np to find the mean and np(1-p) to find the variance, where n is the dataset size and p the probability of a randomly selected student who had paid for coaching. We can then apply a normal distribution to determine the confidence interval (CI).

n=427+2733=3160, p=427/3160=0.1351 approx. Mean, μ=427, σ^{2}=np(1-p)=369.3, standard deviation, σ=19.22 approx. The critical value for 95% CI is 5% (0.025) two-tailed significance. The Z-score for 95% is 1.96. To find the upper bound X, we use (X-μ)/σ=Z, X=μ+σZ=427+19.22×1.96=464.67 approx.

Upper bound is 464 (the largest integer lower than 464.67).