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Math Help - approximations of distributions 2

  1. #1
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    approximations of distributions 2

    A given honest is launched 100 times in a row. What is the probability that in 18 or more of these releases occur side 2?

    I find:
    p = 1/6

    Z = \frac{17,5 - (100* \frac{1}{6})}{\sqrt{100 * \frac{1}{6} * \frac{5}{6}}} = 0,22 -> 0,587064

    But the answer is: 0,4129
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  2. #2
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    Which side of 17,5 do you want? 1 - 0,587064 = 0,412936 That looks pretty close.
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  3. #3
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    Quote Originally Posted by Apprentice123 View Post
    A given honest is launched 100 times in a row. What is the probability that in 18 or more of these releases occur side 2?

    I find:
    p = 1/6

    Z = \frac{17,5 - (100* \frac{1}{6})}{\sqrt{100 * \frac{1}{6} * \frac{5}{6}}} = 0,22 -> 0,587064

    But the answer is: 0,4129
    You have calculated the probability of fewer than 18 not 18 or more.

    CB
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  4. #4
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    Not is P(X >= x) = (P X >= x-0,5) ???
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Apprentice123 View Post
    Not is P(X >= x) = (P X >= x-0,5) ???
    What does that mean? (I'm betting whatever you intend it to mean the answer is no)

    P(X\ge x)=1-P(X<x)

    CB
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