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~Epohes BBBask-Stolz R00B statlonary stote has decided to accept arge shipment ball-polnt pens an Inspection 19 randomly selected pens yields no more than two defec...

Question

~Epohes BBBask-Stolz R00B statlonary stote has decided to accept arge shipment ball-polnt pens an Inspection 19 randomly selected pens yields no more than two defective pens. Find the probability that this shipment accepted 109 of the total shipment dctective. (Use decima placcs )(b) Find the probability that thls shipment not accepted 1590 of the toral shipment defective _ (Use decimal places:)YourKonDLAEboniRlnmutltnOr AMraSubunit Aseignmant

~Epohes BBBask-Stolz R00B statlonary stote has decided to accept arge shipment ball-polnt pens an Inspection 19 randomly selected pens yields no more than two defective pens. Find the probability that this shipment accepted 109 of the total shipment dctective. (Use decima placcs ) (b) Find the probability that thls shipment not accepted 1590 of the toral shipment defective _ (Use decimal places:) YourKon DLAEboni RlnmutltnOr AMra Subunit Aseignmant



Answers

A shipment of 10 items has two defective and eight nondefective items. In the inspection of the shipment, a sample of items will be selected and tested. If a defective item is found, the shipment of 10 items will be rejected. a. If a sample of three items is selected, what is the probability that the shipment will be rejected? b. If a sample of four items is selected, what is the probability that the shipment will be rejected? c. If a sample of five items is selected, what is the probability that the shipment will be rejected? d. If management would like a .90 probability of rejecting a shipment with two defective and eight nondefective items, how large a sample would you recommend?

This problem, we are told that we have a box of 25 and we're taking a sample of three of them to test for defective today. We want to find the probability that a box containing three defectives will be shipped. And so that means we have three defectives And 22 that are not defectives. Now in order for it to be shipped. We don't want to find any defectives at all. And so that means when we choose three of them, we want to choose only from the 22 that are not defective. So 20 to 20 to choose three is the number of different ways in order for that to happen. And then we're gonna divide that by 25 shoes three. Because that's the total number of different ways that we can choose three items from. The total of 25. And so we evaluate this probability And this gives us .66, I'm sorry. .6 696 You and me. We want to find the probability that a box containing only one defective will be sent back for screaming. This this means we have 24 good ones and one defect. We want to find the probability that we find at least one defective. Well, there's your compliment. All this is April 1- the probability that we don't find it any of the defectives. So this is one 24-3 Over 25, choose three For the 24-3. That's the number of different ways that we can choose only good ones From the 24 that we're told there did. And then the 25 choose three is the number of total ways. And evaluating this gives us .12, exactly.

In question 12, were in Chapter six the binomial probability distribution, and then section three areas under any normal curve. Now, the problem that we have says, assume that X has a normal distribution What they specified mean and standard deviation tells us that the mean is 100 and the standard deviation is 15. We're asked to find the probability that X is greater than or equal to 120. So I want to draw this curve, so we get a good idea what we're looking at mean is 100 centre deviation is 15. So that's essentially our spacer, a few standard deviations on each side And on the mark, this value at 120. And shade appropriately, it says find the probably the X is greater than that 1 20 marks. So I wanna shade everything over here to the right. So as I do this, I can use my to 84 calculator and I can use my normal CDF command. And it's gonna ask me for my lower regional. My lower region is where I start shading. Uh that is at 120 and my upper region is where I stop. Well, technically this goes on forever. This curve does. So we're gonna plug in a series of nines to indicate infinity. The mean here is 100 And the standard deviation is 15. We plug this in, we get an area of zero 0912.

So we have a sample of 100 chips and we have 20 that are bad and we're selecting two at a time or ask what is the probability that the second ship that we choose is bad. So the probability is that this can happen is that we have, we choose the first one and that's bad. And we also choose the 2nd 1 which has to be bad by default. And then there's a probability that the first one we choose isn't bad and then the second one that we do choose is bad. Mhm. And this is gonna turn allowed to be 20%. And then we're asked, what is the probability that three year selected at random? What is the probability that they all are defective? Um Since this is just a simple probability, um We're choosing 3, 1, 2 and three. It's just simply going to be the probability that we choose three in a row. That are bad. So we start with 20 of the 100 and then we have 19 bad ones out of 99 And then we go down to 18 bad ones out of 98. And this turns out to be 00705. Yeah. So pretty unlikely. But it could happen.

We want to make a history Graham to begin with and I'm not going to get the whole history RAM but I'll get it set up and we know that we have eight different syringes with the probability of defect being point to one and the probability of accepting a lot or a shipment I guess is supposed to be if you have uh uh less than or equal to one. That is bad. And I put all my data enlist one from zero all the way up to eight. And then in list too I used Binomial pdf yeah eight point on one. Probably they have a defect and then list one and then beautifully it just goes through and gives you all those results. And the gift should give you 10.92274 For a probability of one defect is 10.7457 and so on. All the way down to very, very close to zero. Then part be asked what is the mean number of defects out of eight? And we know that sometimes P which is eight times 80.1 which is 0.8 Part C. We want to find what's the probability that we're going to accept a shipment except a lot. And that means we have to find the probability of zero or one being defective. If that happens that would be accepted. So we need to add these two values together and they add up to approximately 20.997 uh huh. Now part D we want to find what the standard deviation is and that's the square root of n times P times one minus P. For a binomial distribution. So that's going to be eight times point oh one times 10.99 And let me double check what that is. So square root of eight times 80.1 times 0.99 And that comes out to be a small number, 0.2814 And I believe that's the whole question other than you have to get all those values down and graph.


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