# Simple Bernoulli trial: prob of pipe failure during inspection

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• 11-28-2017, 07:29 PM
bulldog160
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.

• 11-29-2017, 07:59 AM
tkhunny
Can you calculate the mean of the given distribution? What is the mean of the Binomial Distribution in terms of it's parameter, p?
• 12-03-2017, 09:53 AM
bulldog160
Quote:

Originally Posted by tkhunny
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.
• 12-03-2017, 02:05 PM
tkhunny
In all situations, the mean is calculated by the definition: $$\sum x_{i}\cdot p\left(x_{i}\right)$$

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?
• 12-03-2017, 06:13 PM
bulldog160
Quote:

Originally Posted by tkhunny
In all situations, the mean is calculated by the definition: $$\sum x_{i}\cdot p\left(x_{i}\right)$$

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.
• 12-03-2017, 07:42 PM
tkhunny
n*p = Mean

n = 23
Mean = ???

Solve for p.
• 12-05-2017, 07:07 AM
bulldog160
Quote:

Originally Posted by tkhunny
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
• 12-05-2017, 08:10 AM
tkhunny
:-) That's where I would start. Good work.
• 12-05-2017, 08:56 AM
bulldog160
Quote:

Originally Posted by tkhunny
:-) 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.
• 12-05-2017, 09:31 AM
tkhunny
Well, that was a Binomial Approximation. Is there a distribution that you feel might be more appropriate? Poisson? Beta? Weibull?
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