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Math Help - [SOLVED] Binomial distribution problem

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    [SOLVED] Binomial distribution problem

    According to Chemical Engineering Progress, approximately 30% of all pipework failures in chemical plants are caused by operator error.

    Suppose, for a particular plant, that out of a random sample of 20 pipework failures, exactly 5 are operational errors. Do you feel that the 30% figure stated above applies to this plant? Comment.

    I did the math and

     p(X=5)= 0.1789

    The book says that this answer means that the 30% figure is acceptable, but I don't understand why. I would have thought that the 30% figure would be acceptable if  p(X=5) was close to 1. Can someone help me understand this? Thanks
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    Probability is a rather elusive subject. Off the bat you have 5/20 = .25 operational failures so this looks good. the probability of this EXACT result is around .18. Think of it this way, we could have 1 operational error, 2,...,20. etc.. and the sum of ALL these possibilities is 1. So when you are saying what is the probability of EXACTLY 5 operational errors. A probability of \approx .18 is definitely one of the highest possible outcomes.

    The key is to understand that there is 20 possible outcomes. If each of the outcomes were eqaully likely, with probability .18. Then the sum of all possibilities would be around 4 and not 1. So the outliers like 1 operational error out of 20 will have a very small probibility, and the out come with the highest probability would be 6 operational errors. Why is this?

    Do you get what I'm saying?
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    I see what you mean about p(X=5) = being the second highest probability and p(X=6) being the highest. I guess my question now is what exactly constitutes the idea that the 30% figure is acceptable? I mean it's the SECOND highest probability, not the highest. To me, because it's not THE highest probability, I would say that the 30% isn't acceptable. It just seems kind of relative.
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    Quote Originally Posted by downthesun01 View Post
    I see what you mean about p(X=5) = being the second highest probability and p(X=6) being the highest. I guess my question now is what exactly constitutes the idea that the 30% figure is acceptable? I mean it's the SECOND highest probability, not the highest. To me, because it's not THE highest probability, I would say that the 30% isn't acceptable. It just seems kind of relative.
    25% operational failures is very close to 30% for such a small sample size. Random process, are just that: random. If you flip a coin 5 times and get 5 heads in a row, you may question the fairness of the coin. The fact is rare events happen, just at a lower frequency. Now, if you flipped the coin 100^100 times you could expect to see heads 50% of the time.

    At this plant we have 5 operational errors when our expectation was 6. Standard deviation takes the relative component out of the equation. I would bet that 5/20 errors is less than one standard deviation below the mean: a very acceptable number. Have you got 75% (average) on every single test you've taken? Or do you deviate, earning +-1 standard dev and the occasional 2.
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    Quote Originally Posted by downthesun01 View Post
    According to Chemical Engineering Progress, approximately 30% of all pipework failures in chemical plants are caused by operator error.

    Suppose, for a particular plant, that out of a random sample of 20 pipework failures, exactly 5 are operational errors. Do you feel that the 30% figure stated above applies to this plant? Comment.

    I did the math and

     p(X=5)= 0.1789

    The book says that this answer means that the 30% figure is acceptable, but I don't understand why. I would have thought that the 30% figure would be acceptable if  p(X=5) was close to 1. Can someone help me understand this? Thanks
    To answer the question you first have to decide on a value such that if the probability is below that value you will reject the claim. A value of 1 is completely unreasonable.
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