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Math Help - Prob that 2 or more people are over 74 inches

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    Prob that 2 or more people are over 74 inches

    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 know that 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!
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    Quote Originally Posted by tbl9301 View Post
    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. Mr F says: Use this information to find the standard deviation. Now read my main post

    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 know that 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!
    Calculate Pr(X > 74) where X follows the given normal distribution. Then calculate Pr(Y > 1) where Y ~ Binomial(n = 100, p = Pr(X > 74)).
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