Thank you for posting your question here at brainly. I think below should be the questions and the answer:
a.Discuss the sampling distribution of the sample proportion
N(p)= 82 >5, n(1-p) = 18 >5, it is normal distribution and provokes central limit theorem
b.What is the probability that the sample proportion is less than 0.80?
z=82-80/sqrt(.82(.18)/100) = .52057
z(x<.52057) = 0.6985
c.What is the probability that the sample proportion is within ± 0.02 of the population proportion?
Sd of error = .0384
+-.02/.0384 = +- .6351
P(z<.6351) – P(z<-.6351)
7357-.2643= 0.4714