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Math Help - flaws in a tape

  1. #1
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    flaws in a tape

    The number of flaws on a magnetic tape produced continuously at a factory follows a Poisson distribution with an average of 0.01 falws per meter. A stand unit of this tape contains 250 meters of magnetic tape.

    (a) What is the probability that there are at least two flaws in a single unit of tape?

    (b) In a random sample of 20 units of tapes, what is the probability that at least two of them are flawless?

    for part a I have 250x0.01=2.5 P(x>=2)=1-P(x=0)-P(x=1)=1-e^{-2.5}*2.5^0/0! -e^-2.5*2.5^1/1!

    for part b, do I just times 2.5 by 20?
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  2. #2
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    Re: flaws in a tape

    You seem to have part (a). Good work.

    Not so good on part (b). First, "times" isn't a verb.
    Second, you now have a binomial problem.

    n = 20

    p = \frac{e^{-2.5}\cdot 2.5^{0}}{0!}
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  3. #3
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    Re: flaws in a tape

    Quote Originally Posted by TKHunny View Post
    You seem to have part (a). Good work.

    Not so good on part (b). First, "times" isn't a verb.
    Second, you now have a binomial problem.

    n = 20



    p = \frac{e^{-2.5}\cdot 2.5^{0}}{0!}
    and what does this mean then, do i use binomial approximation to possion?
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  4. #4
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    Re: flaws in a tape

    Don't need the approximation. You just need n=20 from the question and p from part a)
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  5. #5
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    Re: flaws in a tape

    Please stop guessing. Go reread the section on Binomial, Poisson, Bernoulli, and other discrete distributions.
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  6. #6
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    Re: flaws in a tape

    Quote Originally Posted by pickslides View Post
    Don't need the approximation. You just need n=20 from the question and p from part a)
    no, since there is 0.01 flaws per meter, then 20 units has 20*250 meter, so n is 5000
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  7. #7
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    Re: flaws in a tape

    As previous posters have said, part (b) is a binomial question.

    Step 1 Find probability that any given sample is flawless

    The number of flaws in a sample of length 250 is X \sim Po(0.01 \times 250). ie X \sim Po(2.50)

    It is flawless if X=0. You can find the probability using the formula for a poisson pmf:

    P(X=0) = \frac{e^{-2.5} \cdot 2.5^0}{0!} =  \frac{e^{-2.5} \cdot 1}{1} = e^{-2.5} \approx 0.0821

    This is the probability TKHunny gave you in post #2.

    Step 2 Answer the question given
    "You take 20 independent samples. Each is flawless with probability e^{-2.5} \approx 0.0821. Find the probability that at least 2 are flawless."

    The number of flawed samples follows a binomial distribution. Y \sim Bi(20,e^{-0.25})

    Evaluate P(Y \geq 2) using the method you have been taught.
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