np(1-p) = .036*745*(1-.036) = 25.9 > 10. Therefore, problem approximately follows a normal distribution.
a) mean = .036*745 = 26.82, standard deviation = sqrt(pn(1-p)) = sqrt(.036*745*(1-.036)) = 5.08. Use TI 84 cumulative distribution function with mean 26.28, sd 5.08, upper bound 1099, lower bound 15) = .938
b) normalcdf(mean 26.82, sd 5.08, ub 1099, lb 30) = .266
c) normalcdf(mean 26.82, sd 5.08, ub 35, lb 25) = .586
d) normalcdf(mean 26.82, sd 5.08, ub 1099, lb 40) = .005