Q: A quality control engineer tested 50 items at random from an entire lot of manufactured items and found 1 defective item from the tests. What is probability that there is at most 5% of the entire items are defective?
A: Let us first lay down some facts and assumptions from the problem statement. The 50 items can be assumed to be sampled with replacement if the total number of items is very large. The QC engineer found 1 defective item from these 50 sample items. The total number of defective items is not known, as also the the total number of manufacturing items.
The absence of information on the total numbers means that we have to extrapolate the 1/50 defective item probability to the population's defective item probability. The extrapolation process implies that we can come up with different estimates for the population's probability for different confidence probabilities. This is the essence of the question on the probability that there is at most 5% of the entire items are defective.
One way to solve this question is by using beta distribution. Beta distribution tells us the probability of probability of getting k defective items from n samples. I wrote an earlier blog on beta distribution if you need to review it.
The data k = 1 and n = 50 select the shape of the beta distribution to have the following expression:
f(p) = 2550 p (1–p)49.
This function is then integrated between the lower limit p = 0 and the upper limit p = 0.05 to yield a probability of 73%. The probability of having at most 5% of the entire items defective is 73%. The limits of the integral come from the statement "... that at most 5% of the entire items are defective". Thus, at the lowest, the probability is 0%, i.e., there is no defective item at all, and at the highest it is 5%.
If we plot the result of the integral as a function of the upper limit, i.e., the maximum defective item probability, we find that the probability reaches practically 100% when the upper limit is 10%. This means that the 1/50 defective probability the QC engineer gets from the tests should give him a 100% confidence that the defective probability for the entire manufactured items is maximum 10%.
Notice that by knowing there is a 2% defective probability from the tests, the QC engineer can make an intelligent statement about the maximum defective probability of 10% for the entire population. Neat, eh? We can use math to extrapolate what we know locally to what can happen globally.
Because the available data are so minimal, we cannot use central limit theorem to get a confidence interval. For instance, there is no standard deviation data from the tests. This solution route is therefore a dead end.
There is another method to estimate the defective probability using hypergeometric distribution, but in this case two batches of tests are required.
This function is then integrated between the lower limit p = 0 and the upper limit p = 0.05 to yield a probability of 73%. The probability of having at most 5% of the entire items defective is 73%. The limits of the integral come from the statement "... that at most 5% of the entire items are defective". Thus, at the lowest, the probability is 0%, i.e., there is no defective item at all, and at the highest it is 5%.
If we plot the result of the integral as a function of the upper limit, i.e., the maximum defective item probability, we find that the probability reaches practically 100% when the upper limit is 10%. This means that the 1/50 defective probability the QC engineer gets from the tests should give him a 100% confidence that the defective probability for the entire manufactured items is maximum 10%.
Notice that by knowing there is a 2% defective probability from the tests, the QC engineer can make an intelligent statement about the maximum defective probability of 10% for the entire population. Neat, eh? We can use math to extrapolate what we know locally to what can happen globally.
Because the available data are so minimal, we cannot use central limit theorem to get a confidence interval. For instance, there is no standard deviation data from the tests. This solution route is therefore a dead end.
There is another method to estimate the defective probability using hypergeometric distribution, but in this case two batches of tests are required.
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