Probabilty of 2 or more people over 74 inches in a group...

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...

Thanks for any help!

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

Now we have (74-70)/1.560 = 2.564

Finally, you do the last step...