Suppose the distribution of height over a large population of individuals is approximately normal. Ten percent of individuals in the population are over 6 feet tall, while the average height is 5 feet 10 inches. What, approximately, is the probability that in a group of 100 people picked at random from this population there will be two or more individuals over 6 feet 2 inches tall?

So, P(x > 72) = 0.1 and mu = 70. And I'm pretty sure the P(two or more people over 74) = 1 - P(0 people over 74) - P(1 person over 74) , but I'm not sure how to get from the 72 probabilities to the 74...

The important words are these, "approximately normal".

The big hint, you have, P(X > 72) = 0.10, but you did not see that this should lead you to observe that for a Normal Distribution \(\displaystyle P(x > z_{0.1}) = 0.1\), leading to \(\displaystyle z_{0.1} = 1.282\), (find it in a table or get a calculator to tell you) and finally: (72-70)/s = 1.282 and s = 1.560