Probability of Overhauling a Machine Unnecessarily

[ptlnguyen]

Homework Statement

A machine will be repaired if it produces 5% defected items and won't need repair if the defected items are less than 1%. 50 samples items are tested. If defected items are greater than 0.02, the machine will be repaired, other wise the machine will continue operating.

Homework Equations

(a) What is the probability that a machine will be overhauled unnecessarily?
(b) What is the probability that a machin in need of overhauling will be left in operation?

The Attempt at a Solution

This is my attemp but not sure if it is correct. Please help

Machine needs to be overhaul if >=5% of items are defected
Since only >=0.02 or 2% defected items and the machine will be overhaul, then the difference is: 5% - 2% = 3% . Therefore, the probability that a machine will be overhauled unnecessarily is .003. Am I correct.
Thank you much

 

[Simon Bridge]

Machine does not need repair if defected items are <0.01.
Therefore, the machine needs repair if defective items are what ?

If 50 items are tested, how many defective items need be present before the machine gets repaired?

What is the probability that the machine will need repair, but the required number of defective items does not appear?

Needing repair and getting repaired are different things - your description has two figures for the rate of defects that trigger an overhaul. Is it 5% or 0.02? (and 0.02 what? Percent, proportion, or number of items?)

I'm reading that the machine needs repair at a 1% defect rate or higher but it only gets repaired at 5% or higher ... with a sample of only 50 units, the resolution is 2% (the 0.02 proportion, which is higher than the minimum defect rate that would indicate "in need of repair").

There must be a probability that none of the 50 units are defective, but the machine still needs repair.

 

[256bits]

You are dealing with selecting a sample from a population.
There are tests in statistcs to determine whether your sample is a fairly good representation of the population, or to determine how far your sample could deviate from the population. I am not sure which one to use here, but you must have taken something such as the chi-squared test and others.

In your case, if your machine has a population run of 50 units and you test all 50 then you know exactly how many are defective. If your machine run is much larger, say 1000 units just as a figure to throw out there, then you use a sample size to estimate good/defective units, and the statistical tests will give you an idea if your sample size is large enough or the if the deviation from good/ defective in the sample can represent the good/defective in the population and with how much of a margin of error.

So in extremes, you could have selected 50 good units but all the other 950 units are defective. Or you could have selected 50 bad units and all the other units are good.
and so on in between these extremes.
You are asked to verify from all the possibilities of good/defective that could occur with selection of the 50 unit sample size, how many times you are going to be right or wrong with repair of the machine.

 

[Simon Bridge]

There's context missing isn't there - the approach to use will depend on the methods just taught.

We could treat it as trying to estimate the overall proportion of defectives the machine would produce from the proportion in a run of 50.
So we are looking for how far the sample proportion will be from the "actual" proportion.

 

[rezeew]

.

Your attempt at a solution is not entirely correct. The 5% and 1% mentioned in the homework statement refer to the threshold for repair, not the percentage of defected items. So if the machine produces more than 5% defected items, it will be repaired, and if it produces less than 1% defected items, it will not be repaired. The question is asking for the probability of the machine being overhauled unnecessarily, which means the probability of it being repaired when it does not actually need to be.

To calculate this probability, we need to consider the probability of the machine producing between 1% and 5% defected items. This can be done using the binomial distribution, with n=50 and p=0.05 (since 5% is the threshold for repair). The probability of exactly 1% defected items is given by:

P(X=1) = (50 choose 1) * (0.05)^1 * (0.95)^49 = 0.0000013

Similarly, the probability of exactly 2% defected items is:

P(X=2) = (50 choose 2) * (0.05)^2 * (0.95)^48 = 0.000011

And so on. We can continue this calculation for all values of X between 1% and 5%, and then add up all these probabilities to get the total probability of the machine producing between 1% and 5% defected items. This will give us the probability of the machine being overhauled unnecessarily, since it will be repaired if it produces any of these percentages of defected items.

To answer part (b) of the question, we need to consider the probability of the machine producing more than 5% defected items, given that it actually needs to be overhauled. This can be calculated by dividing the probability of the machine producing more than 5% defected items (which we can calculate using the binomial distribution again) by the total probability of the machine needing to be overhauled (which we calculated in part (a)).

I hope this helps. Remember, probability questions can be tricky, so it's always a good idea to double check your calculations and make sure you understand the question before providing an answer.