In a random sanple of 500 students, 80% say they have a part-time job. An approximate 95% confidence interval for the proportion of students with part-time jobs is?

1. (0.758;0.842)
2. (0.765;0.835)
3. (0.770;0.830)
4. (0.800;0.835)
5. (0.782;0.819)

500 is a fairly big number, so you can use a Normal approximation to the binomial distribution to work this out. The binomial parameters you want are N = 500 and P = 0.8. The mean is just N.P = 400. The standard deviation is sqrt(N.P.(1-P))=8.94. The 95% cutoff value from a Normal distribution is 1.96, so you want the mean +/- (1.96 times the standard deviation), both expressed as proportions of 500. Try it out, and see if you get any of the answers suggested.