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Math Help - AP stats

  1. #1
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    AP stats

    About 60% of customers at a restaurant want to sit in a non-smoking section. A new restaurant with 120 seats is being planned. How many seats should be in the non-smoking area in order to be very sure of having enough seating?

    How would I find this?
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  2. #2
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    Quote Originally Posted by Morgan82 View Post
    About 60% of customers at a restaurant want to sit in a non-smoking section. A new restaurant with 120 seats is being planned. How many seats should be in the non-smoking area in order to be very sure of having enough seating?

    How would I find this?
    The simple-minded approach would be to calculate the expected number of non-smokers: (n)(p) = (120)(0.6) = 72.

    But how sure is 'very sure'? 99% sure? Having 72 seats wouldn't make you 99% sure ...... (in fact, it would only make you about 53% sure!).

    Let X be the random variable number of non-smoking seats required.

    Then X ~ Binomial (n = 120, p = 0.6)

    To be say, 99% sure, you need to find the smallest value of x such that \Pr(X \leq x) > 0.99. This value of x is the number of seats required.

    You can use technology to very easily solve \Pr(X \leq x) > 0.99 for x ..... I get x = 84.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post

    You can use technology to very easily solve \Pr(X \leq x) > 0.99 for x ..... I get x = 84.
    Technology Assistance for Morgan82:

    Binomial Distribution

    For a 95% confidence ratio, use x = 81
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