We take a simple random sample of 100 people and check their ages. Out of the 100, 15 are 65. What is the probability that the proportion of people over 65 in the entire population is between 13% and 17%? What error range can we guarantee with 90% certainty?

Answer 1

Use the normal approximation to the binomial distribution.

mean = np
sd = √npq (where q = 1-p)

Therefore:
mean = 100 * .15 = 15
sd = √npq = √(100 * .15 * .85) = 3.57

Use z-scores:
z = (13 - 15)/3.57 = -0.56
z = (17 - 15)/3.57 = 0.56

Use a z-table to find the probability between the two z-scores.

For the second part, use a 90% confidence interval formula.

I hope this will help get you started.