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Thread: [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

    $\displaystyle p(X=5)= 0.1789$

    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 $\displaystyle p(X=5)$ 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 $\displaystyle 5/20 = .25$ operational failures so this looks good. the probability of this EXACT result is around $\displaystyle .18. $ Think of it this way, we could have $\displaystyle 1 $ operational error, $\displaystyle 2,...,20.$ etc.. and the sum of ALL these possibilities is $\displaystyle 1.$ So when you are saying what is the probability of EXACTLY $\displaystyle 5$ operational errors. A probability of $\displaystyle \approx .18$ is definitely one of the highest possible outcomes.

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

    $\displaystyle p(X=5)= 0.1789$

    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 $\displaystyle p(X=5)$ 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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