Bernoulli's theorem, probability

Question

In an exam, candidates fail with a chance of p = 0.6, (and pass with a chance of q = 1 - p = 0.4).

A total of 90 are taking the exam.

We would like to estimate p based on Bernoulli's theorem.

What is the probability that the difference between the estimated and the actual value will be less than 0.1?

Answer 1

90p = 90(.6) = 54 fail, 36 pass on average, with a standard deviation of the sqr(npq) = sqr(21.6) =4.65

95% of the time, the failures of 90 students would be 45<F<61

But with Bernouilli distribution, or a binomial expansion P(F=54) = 90C54 times (.6)^54 times (.4)^36 which should be a relatively small probability. calculate 17 different probabilities and sum them P(F=45)+P(F=46)+ ....+P(F=60)+P(F=61) and it should be closer to the 95% probability that failures of 90 students is within 45 to 61. Or p=0.95 = P(45<F<61). p=.6 would be less than one standard deviation from the mean or calculate 9 probabilities and sum them to get P(50<F<58) = P(50)+P(51)+....+P(57)+P(58) = 90C50(.6^50)(.4^40)+ ....