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5. (3 + 3 points) During a quality check on a batch of mobile phones, a double is performed sampling plan as follows. First; a sample of 50 mobile phones is randoml...

Question

5. (3 + 3 points) During a quality check on a batch of mobile phones, a double is performed sampling plan as follows. First; a sample of 50 mobile phones is randomly selected. lemons for control. If all of these are correct (not defective) , the party is accepted, but if more than or equal t0 two are defective, the Iot is rejected. If exactly one is defective, proceed to a second selection where another 50 mobile phones are checked: If all of these are correct, the party is accepted, otherwise r

5. (3 + 3 points) During a quality check on a batch of mobile phones, a double is performed sampling plan as follows. First; a sample of 50 mobile phones is randomly selected. lemons for control. If all of these are correct (not defective) , the party is accepted, but if more than or equal t0 two are defective, the Iot is rejected. If exactly one is defective, proceed to a second selection where another 50 mobile phones are checked: If all of these are correct, the party is accepted, otherwise rejected the. Assume that the error rate in the batch is p 2% defective (a) Assume that the batch is very large so that the sample sizes are negligible compared to with the size of the party. Calculate the probability that the party will be accepted. (b) Assume now instead that the size of the lot is 300. Give an exact expression of the probability that the party is accepted. The term is allowed t0 contain binomial coefficients which does not need to be calculated.



Answers

A very large batch of components has arrived at a distributor. The batch can be characterized as acceptable only if the proportion of defective components is at most. 10. The distributor decides to randomly select 10 components and to accept the batch only if the number of defective components in the sample is at most 2.
(a) What is the probability that the batch will be accepted when the actual proportion of defectives is .01$? .05 ? .10 ? .20 ? .25 ?$
(b) Let $p$ denote the actual proportion of defectives in the batch. A graph of $P($ batch is accepted) as a function of $p,$ with $p$ on the horizontal axis and $P($ batch is accepted) on the vertical axis, is called the operating characteristic curve for the acceptance sampling plan. Use the results of part (a) to sketch this curve for $0 \leq p \leq 1$ .
(c) Repeat parts (a) and (b) with "I" replacing "2" in the acceptance sampling plan.
(d) Repeat parts (a) and (b) with "15" replacing "10" in the acceptance sampling plan.
(e) Which of the three sampling plans, that of part (a), (c), or (d), appears most satisfactory, and why?

Okay. So we got a little hyper geometric distribution going on here, which is similar to a binomial distribution at small sample sizes. And in part a were asked to identify with the big end the population sizes and they told us that there is a shipment of 250 so that's gonna be it. And from those 250 we took out four fuses Of which there is a 96% or 94% chance that they work. So I'll be the peak. So that concludes part A. Now we go to part B where we apply the formula and it's a pretty lengthy one. Uh So I'm just gonna not right all this out. Just go ahead and right on papers if you get the same answer as me. Okay. And as I was writing this on paper, it occurred to me that we did not have an X. So we're just gonna leave those exes there. This is the answer for a beat. Okay. Yeah. And part C. And this right here is going to be the binomial distribution since our sample size is less than 5% of the total population Of 250 up here. Um and now we just need to film and the I filled in the probabilities which are .94.06 and now we just need to fill in what the end is and that we are given is for she was going to go ahead and replace these ends with force like this. And that is going to be the binomial distribution. Then just as a comparison for party, we can observe that This is 1- the probability of the fuse being good and then this is the probability that refuses bad. Similarly, this is all the fuses that are expected to be good times. All the fuses cider can be bad, divided by a pretty big number and this is just to get it back up there. So these should be approximately the same result

28 question a probability off three is equal to 13 c three to C 0/15 C three, which is 0, 00.63 authority off two is equal to 13 c two to see 1/15 C three She's ableto open people. We shall see probability off one is equal to 13, 81 times To see to over 15. 33 years is a point or 29

Here we have 12 refrigerators, seven of which have a defective compressor, and the other five do not. From the 12 we're going to randomly select six, and we're interested in the number of this election that have a defective compressor. So this is a hyper geometric random variable, the number that do you have a defective compressor because we have a population with two distinguishable groups that form of partition, and we're sampling six from that population without replacement. So the number X, which belongs to one of those groups, is a hyper geometric, random variable. So, as stated in the problem, we have a population of refrigerators, which equals 12. And in the group that we're interested in, that's the seven that have defective compressors that size of seven, and we're looking at six of them, so the selection sizes six or the sample size is six. So in party were asked for the probability that five of them of the sixth sampled have a defective compressor for hyper geometric random variable for the probability that five given a sample size of six, a group size of seven and a population size of 12. So before making this calculation. I'll just write out the general formula. So the probability mass function. So it's equal to amateurs X times and minus M over a small and minus X divided by population size. Choose the sample size so that corresponds to the following in our situation, and this should actually be choose. And this comes out to zero point 114 next for Part B were asked for the probability that at most, four have a defective compressor from our sample of six. So this is equal to this summation and notice that I'm summing Starting at X equals one so that we can't have X equals zero. Because if we're sampling six from a population of 12 then and seven of the 12 have a defective compressor, then at least one of the sampled refrigerators must have the defective compressor that's given by the following requirement for the number for X have the max of zero and n minus. N plus M is listener equal to X, which is Lessner equal to the minimum of small end. And, um so so that is, if X is a hyper geometric random variable, its value must be an integer that which satisfies this expression. Doing this calculation comes out to zero point 879 in Part C. We're looking for the probability that exceeds its mean value by more than one standard deviation. So we have to calculate both the mean value and the standard deviation for this random variable for hyper hyper geometric random variable. The expectation is given by this formula, which comes out to 3.5, and the variance is given by this formula. So we have 12 minus 6/12, minus one times six times 7/12 times one minus 7/12, which comes out to 0.795 which gives us a standard deviation, which is the square root of that value of 0.892 So we're looking for the probability that X is greater than the mean plus one standard deviation. So that's the probability that is greater than the mean, which is 3.5 plus the standard deviation 0.892 which is the probability that that's greater than 4.392 And since X can only take on your values, this is the probability that it is greater than five greater than or equal to five. And since our sample size is six, then this must be equal to the probability that X is equal to five, plus the probability that it is equal to six. This is actually the complement of what we calculated in Part B, and this is 0.1 to 1. And then, lastly, for Part D, were asked to assume that there are 400 refrigerators, 40 of which have defective compressors and were randomly selecting 15 of them. So whereas to approximate the probability that at most, five of that selection have defective compressors and we're as to provide, um, or convenient or less tedious way to calculate this than to use the hyper geometric distribution. So this is a situation where you can use the binomial distribution to approximate ah, hyper geometric. And this is admissible because we have a large number of population, so 400 and the possible number of successes is high, which is 40. That's the number of defective compressors among the 400 refrigerators. So if you have a large population and a large number of successes, which is M, you can use the binomial approximation. And to use the binomial approximation, this probability success is equal to am over the population size. So that's 40 over 400 in this situation or a 0.1. So to find the probability that at most five from our random selection have a faulty compressor. This is approximately the cumulative probability for a binomial based on a sample size of 15 and a probability success of 40 over 400 or 0.1. And this 15 is the size of our selection for Part D. So out of the 400 refrigerators, we're selecting a random sample of 15, and we're interesting the probability that less than five of these 15 have defective compressors, and this probability comes out to zero point 998

Problem. 35. We have a computer ship that contains 10,000 computer ships. Then the population equals 10,000. Compute computer ships. Unfortunately, 50 of them are defective. We have 50 effective. If it's still chips party, we want to compute the probability that two randomly selected ships then some build equals two. These two will be defective using conditional probability. Then for birthday, we want to calculate the probability of two of them. Let's see, do not. The first event is mhm. The first event to be defective is E and intersection. By the second event is it then e represents the first sambal. The first take in computer ship would be defective. If represents, the second computer ship will be defective. The intersection of them using conditional probability equals the probability four, the second for the first to be defective was employed by the probability of the second to be effective. This is the first and this is availability for the second to be detective, Given that the first is different, this is the way that we write conditional probability. Then we will calculate the probability of being effective for the first, Just the number Oh, the effect of ships divided by the total number of the population. It's 50 divided by 10,000, multiplied by the probability for the second event to be defective. Given that the first is defective after we sell selecting a defective ship because we know now that the first is defective, this means we have only 49 effective remaining, and the population will be produced by one. Then it's a 50 minus one, divided by 10,000 minus one bike and creating this 50 market by by 49 divided by 10 thousands. But the boy boy 9999 it equals 2.45 the blood by 10 to a lot of negative five, and we can write it in percentage. It equals Oh point oh 245%. This is a percentage to have both of them defect for birdie. Given that the probability that the first ship randomly selected is 50 divided by 10,000, which is calculated here, we want to compute the probability that two randomly selected ships are defective under the assumption of independent events. Then we would assume and dependent events. Why do we assume that and It's not true because we have seen that the second event is dependent on the first because the first is defective. We have reduced from here one. If it's not defective, we will not reduce one. Which means it depends on the first event. Then, if depends on mm. But now we will assume independent events and we will calculate again the probability of E the sector equals the probability of e multiplied by the probability of equals 50 divided by 10,000, multiplied by the probability of the second will be defective as an independent event which means will assume that that somebody is without is with the replacement. Some bling is with replacement. Then the probability of F equals the probability of 50 divided by 10. Some of them calculating this probability. Gifts 2.5 multiplied by 10 to the bottom of negative five and we can write it in percentage equals opened. Oh, 25%. We can see that the two values are very close to each other, which means we can use with this assumption four small symbols. We can notice that we have taken only two samples out of the 10,000, then some billing percentage equals two equals two divided by 10,000 or the symbol relative to the population. Size equals only all 0.2% and as long as this value is smaller than 5% as a rule of thumb, then we can use the assumption, oh, independent events.


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