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The Acme Pro Office eNce provides temporary office personnel for many corporations_ They have found that 88%6 of their invoices are paid within 10 working days by t...

Question

The Acme Pro Office eNce provides temporary office personnel for many corporations_ They have found that 88%6 of their invoices are paid within 10 working days by these corporations Suppose the service randomly selects their invoices that are at least 10 working days old Use this situation to answer the parts below Recognizing that this is binomial situation, give the meaning S and in this context. That is, define what you will classify as success and what you will classify as "failure refe

The Acme Pro Office eNce provides temporary office personnel for many corporations_ They have found that 88%6 of their invoices are paid within 10 working days by these corporations Suppose the service randomly selects their invoices that are at least 10 working days old Use this situation to answer the parts below Recognizing that this is binomial situation, give the meaning S and in this context. That is, define what you will classify as success and what you will classify as "failure refers to paying invoices within 10 working days_ Next , give the values of and q . (Recall that is the probability of failure in trial: Construct the complete binomial probability distribution for this situation in table out to the right Using your table, find the probability that exactly three of the chosen corporations paid within 10 working days Find the probability that at least six paid within 10 working days Find the probability that at most paid within 10 working days Find the mean and standard deviation of this binomial probability distribution. By writing sentence, interpret the meaning of the mean value found in (g) as tied to the context of the Acme temp service and paid invoices unusual to have only invoices to have been paid within 10 days? Briefly explain your answer giving supporting numerical evidence:



Answers

For each scenario (a)-(j), state whether or not the binomial distribution is a reasonable model for the random variable and why. State any assumptions you make. (a) A production process produces thousands of temperature transducers. Let $X$ denote the number of nonconforming transducers in a sample of size 30 selected at random from the process. (b) From a batch of 50 temperature transducers, a sample of size 30 is selected without replacement. Let $X$ denote the number of nonconforming transducers in the sample. (c) Four identical electronic components are wired to a controller that can switch from a failed component to one of the remaining spares. Let $X$ denote the number of components that have failed after a specified period of operation. (d) Let $X$ denote the number of accidents that occur along the federal highways in Arizona during a one-month period. (e) Let $X$ denote the number of correct answers by a student taking a multiple-choice exam in which a student can eliminate some of the choices as being incorrect in some questions and all of the incorrect choices in other questions. (f) Defects occur randomly over the surface of a semiconductor chip. However, only $80 \%$ of defects can be found by testing. A sample of 40 chips with one defect each is tested. Let $X$ denote the number of chips in which the test finds a defect. (g) Reconsider the situation in part (f). Now suppose that the sample of 40 chips consists of chips with 1 and with 0 defects. (h) A filling operation attempts to fill detergent packages to the advertised weight. Let $X$ denote the number of detergent packages that are underfilled. (i) Errors in a digital communication channel occur in bursts that affect several consecutive bits. Let $X$ denote the number of bits in error in a transmission of 100,000 bits. (j) Let $X$ denote the number of surface flaws in a large coil of galvanized steel.

Now itself exercise to the point of 19. So for each day valves, three vendors will get the people order and has There are two days and the total situation is not vehicles three times three. Here. The first letter denotes a vendor who gets the order on the first day, and the second letter denotes a vendor who gets order on the second day. For example, Way to waste remains that the second vendor gets the order on the first day and the third vendor gets the order on the second day for Part B. Since the orders are selected randomly. Hence, it's reasonable to assign the same probability to each of these simple points. And hence their probability is 19 for each Percy. The Invent A is the event that the same vendor gets both order, so there are only three cases well, when they want to do it too obvious. Three. Mystery and Event B is an event that the two gets at least one order, and you can see that there are five regulations. Mhm results with everyone with a mystery with revision of history, history and the other five situations all satisfies the condition of inventing, okay? And also know that this is a simple point, which means they are exclusive. And for each result, only one of the nine simple points can have happen. Enhance according to X m three of definition 2.6. Yeah, probability of inventing a is just three overnight, which is also one third. And the probability of B is 59 and a set of the zoning of a entity. I know that here we should add ribbon movement V three V three to be, and hence their seven situations in the union and the problem. The probability is 7/9, and for the intersection of A and B here, you can see that owning the element V two V two is in both events and has it is the only situation of the intersection. The probability of the intersection is 19

Yeah. There are nice sample points in this experiment. Since for the first day there are three possibilities. And for the second day there are three possibilities as well. So three times 3 calls nine, we assume that the vendors are selected at random each day. So we assign the same probability to each sample point. That is the probability of each sample point is one night. A. Is the event that the same vendor gets both orders and hence A. Includes. They won they won the two v. 2 and the history the history P. Is the events that we to get the next one order. So be includes. We wanted to which you revenge We too. We too. Way to A. three. And this really to So the probability A is just the sum of the probability of all the sample points it includes. So it equals 1/3. The probability of B is also the some of the probability of all the simple events it includes. So the probability will be is 5/9. Know that a union be includes. We won we won. We've only two way to live on A two, a 2 Page two of the three The three v. 2 and This revision is three and a half. The probability is 7/9. And a intersect B already includes. Way too. Way too Enhance its probability as one or not.

Were given that delightful has a 0.2 chance of failing, and each box has ate light bulbs in it. So the first thing we're gonna do is find the probability distribution. Now, on this page, on the last page of the book, you were given a lot of notes about how to do these problems on a computer. That's because they want you to do these problems on either a computer or graphing calculator. I'm gonna be doing this and excel well, you conduce this in over programs. The steps might be a little bit different, though, So if we do this in Excel, first thing I'm gonna do is make a list of all the possible outcomes. There's eight trials, so we need the number 0 to 8. Just this select all home and dragged down. They don't want to going to be one under formulas and statistical, and I use the binomial distribution. Tual. So our trials that's eight. Cause there's ate light bulbs in the box or probability was 80.2 We don't have to convert it like we normally deal because we weren't given it as a percentage. It's already in the form we wanted to be in Cumulative is false for not working on this right now. So for this the number I'm gonna play it all my numbers from 0 to 8. So starting with B a one box colon, the A nine rocks, and we get this. So now we need to draw a hist. A gram of acsa. So for a history Graham of acts, we put everything in order, and our X values are going to be the X access. So that's just gonna be the numbers from 0 to 8. So to free 456 seven and eight. No wonder why access is gonna be all of the probabilities. However, what? Let's go back to take a look here. You notice the 1st 1 is the biggest. Then the 2nd 1 smaller, smaller, smaller and keeps getting smaller and smaller. This e means times 10 to the blank power. So, for example, on the probability of seven bull to being bad is one times 10 to the negative, a lower power. So the reason why it keeps getting smaller is because of this probability is so low that getting zero, where wine that's the most realistic outcome so we can make our hissed a gram like a descending staircase. Kind of. So zero is gonna be the highest, then wine and to than free four and eventually these ones, they're going to get so small that it's just a tiny little dash. And then we'll fill in some values with one for zero. That was a point. Eat by one one at one that was 10.139 Then we have points cereal 099 I don't put one more. The one at free. We have 10.0 for. So this is Ah, a decent history, Graham. So what is the probability of getting a box with no failing light bulbs? So let's go back to our distribution that we just made it. So no feeling like bulbs that zero. So that's right here. The 0.851 This problem is just reading the table. So 0.851 That's because the no here means zero. What is the probability of getting a box with no more than one? So what we'll do for this one is will take all the possible options. No more than one means even zero or one. So we'll take both of these two and add them together. And when you do that, you should get a point 990 We just add the 1st 2 together, find the mean and standard deviation of X. So because we have a binomial, random variable over the light bulb fails or it works. So it's binomial. These are the two equations we used to get mean and standard deviation. Your notes might say a que here instead of one minus p. That's the same thing. So N is our number of trials, that is eight and P is our probability. That's 80.2 so are mean is equal. Chill eight times 80.0 chill, which is equal 2.16 Our standard deviation is the square root of eight times 80.0 chill times one minus 10.2 or 0.98 If you're using keel, it's the same feign, and this gives us point free 96 So what proportion of the distribution is between you have these two symbols. So first, let's define them. This one is called new, and it's the mean. This one is called Sigma, and it's the standard deviation. So this is asking how much of the data is between mean minus standard deviation and mean plus standard deviation. Well, mean minus dinner deviation. That gives us a negative. Zero is our lowest option. So we have starting from zero chill. But what's 0.16 plus 0.396 Find six. Well, that's Ah, 0.5 by six. So we go here if you notice we have all whole numbers here. So 0.556 That doesn't get us toe one because this is binomial. We can't have half of a try algo ever way. We can't have a light bulb half working, so we have to round everything down and we're only gonna be in the zero box, for example. Let's say we got a mu plus Sigma was equal to free. Then we would include zero to free, but because it hasn't hit one, we're going to stay in the zero box. So we're gonna get 0.851 again. Now, what about if we multiply the standard deviation by two first? So we're still gonna have zero here? It doesn't matter what the exact value is its negative. So we're still gonna have zero, and we're gonna get 00.952 here again. It didn't hit one yet, so we're still within that 85% of data or 0.851 So how does this information relate to the empirical rule and Chevy sex? Fear out. These two rules give less guidelines for how much of the data is going to be within one standard deviation of the mean, like in part E two standard deviations of the mean in free standard deviations of the mean. And sure enough, most of the data should be off in one standard deviation last living too and less within free. However, in our problem, because the probability of a light bulb not working is so low. Almost all of the data is within one standard deviation and so and relates to this because this is what we would expect from such a low probability. Okay, if we buy 100 box isn't we're going to simulate how many of them were go bad. There are many different ways of doing random numbers simulations. The way I'm going to do it is just one way of doing it. So what? I'm gonna do is generate ah 100 random numbers ranging from 1 to 100. I'll do this by going to insert function under category all, and I'm going to scroll down to the ours and almost the aerial ran between. I'm gonna generate numbers from 1 to 100. Now, what my role is going to be is any time we see the number one or two, that means it is going to be a broken bull. That's because we have a 20.2 chance of a light bulb failing. So that's really 2%. So there's 100 boxes, 2%. So wanting to that's 2% of 100 because that's two numbers. Now, we're going to do this 100 times, so I'm going to drag this all the way down to Ah 100. I'm gonna quickly look through and see if we got any ones or two. I see it too. I see one. Okay, so we got to Obviously, your answer could realistically be 0123 or four. Now, repeat, part h in common on it. So we're gonna do this a few times now. We're gonna comment on what we see happening, Nina. So it's high, like this whole thing. This is why I'm doing it this way. So is gonna be very easy to recreate, because I just drag this to the right. I'll do free more. Okay, so starting at the 2nd 1 I see a to there's a one. A number one so free. So we got to for the 1st 1 Been free. Now let's look at this line. No. Is that looks like zero must quickly make sure Yep, zero. And then the last one. There's a one. There's an over one. So, too, to so free we got free. This time you notice we're kind of getting around the same amount each time to or free makes sense. There's a 2% chance that a bold goes bad, so the higher this number gets, we should be getting closer to 2% each time. So 12 or free

So we're gonna be taking a look again at the lottery from Arizona. The probability of winning is you gotta have at least one number right? Which is going to be given by one minus the probability of getting none of the numbers, right? So this right here is p f zero 0.373 is the probability of getting zero numbers. And when we subtract what we get is 0.6 to 87% chance are points 62.87 chance percent chance of success. So will this is equal to P and we can plug that in there for the next part, which is just part A. So we'll just write it like this P of big X is equal to the little X. Mhm of 0.6 to 87 times point 3713 Since one minus 10.6 to 87 has to be 870.3713 The power of x minus one. And that is part A right there, so not for part B. The probability that it takes you exactly three weeks is just going to be a matter of substituting this X with a three uh right there and there. So what we get as a result is 0.867 So that's going to be part B. The first part and the second part they're asking for the probability of number of trials taking at most three. So X is going to be less than or equal to three and this is going to be given by this formula Mosul, that would be one minus the probability of X equaling four. Yeah. And then we can solve just like for this again, instead of putting the three up here, we're gonna put it for. So this part right here, it's going to be points 6 to 87 times 870.3713 to the power of four minus ones three. Yeah and this is equal two point oh 3 to 2 and then we just subtract one for that. So we get yeah one minus 10.322 or 0.9 six seven eight. Yeah it's an interesting number. And then at least three that is going to be P. X. Is at least three. So it can be equal to three. That is going to be the probability that it is that X is three plus the probability that it's four and so on. Yeah. Mhm. Until we get like an infinite number just so really high on. So we'll just call that to sell and this is going to be about equal to 0.3


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