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Thread: Simple Bernoulli trial: prob of pipe failure during inspection

  1. #1

    Simple Bernoulli trial: prob of pipe failure during inspection

    I'm really getting baffled with this question that has taken me far too long to complete and would love some guidance.


    An accident caused the catastrophic failure of metal pipes in a factory. There were
    six metal pipes in the garage at any given time.

    Table 1 shows the numbers of metal pipe failures that had occurred on
    each of the 23 previous inspections.

    Table 1 Number of metal pipe failures

    Number of failed metal pipes 0 1 2 3 4 5 6
    Number of inspections------16 5 2 0 0 0 0


    (i) Let p be the probability that a pipe fails on an inspection. What
    distribution is appropriate to describe the failure or non-failure of a
    particular metal pipe on a particular inspection?

    For this I have said that this is a Bernoulli Distribution due to only two possible outcomes of failure and non-failure.

    (ii) A reasonable estimate of p is 3/46 or 0.065. Explain where this number comes from.

    This is where I am getting stuck on. I cannot work out what p is using the information they have provided.


    Many thanks for your help.

  2. #2
    Elite Member
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    Can you calculate the mean of the given distribution? What is the mean of the Binomial Distribution in terms of it's parameter, p?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    Quote Originally Posted by tkhunny View Post
    Can you calculate the mean of the given distribution? What is the mean of the Binomial Distribution in terms of it's parameter, p?
    The mean of a binomial distribution is np, but I don't know how to calculate the p in this particular question.

  4. #4
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    In all situations, the mean is calculated by the definition: [tex]\sum x_{i}\cdot p\left(x_{i}\right)[/tex]

    In your case, you have 23 inspections. Using the formula, above, we have: 0*(16/23) + 1*(5/23) + 2*(2/23)

    There is your Mean. Now what?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  5. #5
    Quote Originally Posted by tkhunny View Post
    In all situations, the mean is calculated by the definition: [tex]\sum x_{i}\cdot p\left(x_{i}\right)[/tex]

    In your case, you have 23 inspections. Using the formula, above, we have: 0*(16/23) + 1*(5/23) + 2*(2/23)

    There is your Mean. Now what?
    I genuinely don't know what to do after this.

  6. #6
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    n*p = Mean

    n = 23
    Mean = ???

    Solve for p.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  7. #7
    Quote Originally Posted by tkhunny View Post
    n*p = Mean

    n = 23
    Mean = ???

    Solve for p.
    If the mean is 9/23, and n=23,

    then p = 9/23 / 23 = 9/529

  8. #8
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    That's where I would start. Good work.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  9. #9
    Quote Originally Posted by tkhunny View Post
    That's where I would start. Good work.
    But the answer for p in the question is 3/46 or 0.065, whereas I got 9/523.

  10. #10
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    Well, that was a Binomial Approximation. Is there a distribution that you feel might be more appropriate? Poisson? Beta? Weibull?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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