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On very hot summer day; 5% of the production employees at Midland States Steel are absent from work. The production employees are randomly selected for special in-d...

Question

On very hot summer day; 5% of the production employees at Midland States Steel are absent from work. The production employees are randomly selected for special in-depth study on absenteeism. What is the probability of randomly selecting 10 production employees on hot summer day and finding that none of them are absent?Multiple Choice0.0020.3440.5990100

On very hot summer day; 5% of the production employees at Midland States Steel are absent from work. The production employees are randomly selected for special in-depth study on absenteeism. What is the probability of randomly selecting 10 production employees on hot summer day and finding that none of them are absent? Multiple Choice 0.002 0.344 0.599 0100



Answers

Assume that Google, Inc. hires employees on the different business days of the week (Monday through Friday) with equal likelihood. a. If two different employees are randomly selected, what is the probability that they were both hired on a Monday? b. If two different employees are randomly selected, what is the probability that they were both hired on the same day of the week? c. What is the probability that 10 people in the same department were all hired on the same day of the week? Is such an event unlikely?

10 employees a randomly selected and will be interviewed about company loyalty. So Part A is asking. Would the selection of the 10 employees be a final meal experiment? And the answer would be yes. And the reason being is a binomial experiment has to meet certain criteria, and the first thing it must meet would be. Are there a fixed number of trials? And in this instance, yes, there are a fixed number of trials because we're going to ask 10 employees about their loyalty. Are they loyal or are they not? The second would be that those trials are independent of each other. And again, the answer would be, yes, these air independent trials. Because when I'm asking these 10 employees, the response of one employee is not dependent on the next employee, and the third thing that must hold true, there can only be two outcomes. And again there are only two outcomes because we're going to ask about their loyalty. Are they loyal? Yes, or are they not loyal? So you're the two Outcomes are yes or no Part B. So we're no, we're running a binomial experiment and we know that we're going to be speaking with 10 individuals. So R N Value is 10, and a Randstad Harris Interactive survey survey reported that 25% of employees said that their company is loyal to them. So our P value is going to be 0.25 and P stands for the probability of success. Our que value represents the probability of failure. So if 25% believe the company is loyal to them, that means there are 75% out there that say no, the company is not loyal. And for part B, we want to know what is the probability that none of those 10 employees will say that their company is loyal to them. So we're talking about an X value of zero now when solving binomial probabilities, the probability of an event will be equivalent to N. C X multiplied by P to the X power multiplied by Q to the n minus X power. So in this particular problem, R N was 10. See, our X was zero, and we're going to multiply that by our probability of success raised to the end or the X power. So in this case is to the zero power and Then we multiply by the Q value to the n minus X power, and that will be the probability of X equaling zero. So at that point, we're going to utilize our calculator. So I'm going to bring in my graphing calculator. So we're going to do 10, and we're going to scoot over to the Probability and C zero. And then we're multiplying that by 0.25 raised to the zero power and multiplying that by 0.75 raised to the 10 minus zero power. And we're going to get a probability of approximately zero five six three. Let's go on to Part C in Part C. We want the probability that four of the 10 employees will say that their company is loyal, so our end value is still 10 RPI value is still 100.25 r Q value is 0.75 The only difference now is that our X value will be four. So the probability that X equals four will now be 10 C four multiplied by 0.25 to the fourth, power multiplied by 0.75 raised to the 10 minus for power. And when we typed that into our calculator. So I'm gonna bring in my calculator again. I want that statement and I want to back up and I want to change this zero to a four. I want to change this to a four, and I want to change this to a four. So now I have 10 C four times 25 2.25 to the fourth Power times 0.75 to the 10 minus four Power and I'm going to get a probability of 0.14 5998 And if I wanted to take that to four decimal places, that would be 40.1460 Now for the final part of this problem, Parts D, we are going to need to construct a probability distribution for the possible X values. So for Part D, we are saying, What's the probability that at least two of the 10 employees will say that their company is loyal to them so n is still 10, but our X values that we are interested in are to three, four, 567 eight, nine and 10 so we can use that formula once with the two for the X ones with the three for the X ones with the fourth for the X and so forth. Or we can let our calculator do a lot of the work for us. So what we're going to do is we're going to construct a chart with all of the's X values, and we're going to let the calculator generate all the probabilities. So here's my calculator. I'm going to hit this stat button and I'm going to clear out any lists before I start, just in case there's anything in them and there happens to be a little bit more. Okay, so we're gonna go back to list one, and we're gonna put in all of the X values that are of interest to us. So 23456789 and 10. So enlist to we're going to tell the calculator to do 10. C x and X is going to change, multiplied by 0.25 to the X power and then multiplied by 0.75 to the 10 minus X power. And the way we're going to put that into our calculator is we're going to type it in as 10. We're gonna access the combination feature. And then instead of typing in X, since we have placed all of our X values in tow list one we're gonna say list one multiplied by r p value raised to. And instead of saying X, all of our X values are enlist ones were going to say, raised to the list one power times 75 raised to the 10 minus X, which is list one power. And now it has generated all of the individual probabilities for each of those X values. I'm gonna ranked about two to death or three decimal places, so I'm gonna have 30.282 point 250.146 point 058 016003 Uh, this one is in scientific notation, which means move the decimal 10.4 places to the left. So we have three preceding zeros and then at three 86 in the probability for X, being nine will have four preceding zeros and then 286 And for 10 we're going to have six proceeding zeros followed by 954 So for us to find the probability that X is at least which is greater than or equal to two. I am going to need to sum up all of the's values, and the fastest way to sum them up is to let the calculator sum up list, too. So I'm going to quit and I'm going to hit Second stat and I'm going to ask the calculator to sum up all the values enlist to, and we will get a probability of 0.75597 So as a to four decimal place answer, it would be 40.7560 So let's quickly summarize all of our answers. Part A. Is this a binomial experiment? Yes, it is because it has met the fixed number of trials. All the trials were independent, and there were only two outcomes for Part B. What's the probability that none of the 10 people believe or say their company is loyal to them? It would be 0.563 for Part C. What's the probability that for of the 10 people will say their company is loyal? The answer would be 0.1460 And for Part D. What's the probability that at least two of the 10 will say that their company is loyal to them? It would be 100.7560

Problem. We have 30 workers in the day shift 22 from the evening and 15 from the morning shift. I want to choose nine workers. So what's the probability that all nine are from the day shift? Ah, well, we have 30 50 to 67 peoples of 67. Choose nine is gonna be the denominator. And if we want them all to be from the day shift, that would be 30 c nine. So when you put that into a calculator, that will work out to, ah zero point 000 335 um, part B. All nine are from the same shift. So that means Ah, nine day shift or nine evening shift or nine morning shift. So we'll have to add a three cases together. That would be 30 C nine over a 67 C nine plus a nine from the evening shift would be 22 c 9/67 c nine plus Ah, nine morning shift would be 15 c 9/67 c nine. And if you add it all up, put in the calculator. That will be 0.346 part C At least two shifts are represented. So Ah, we could have two represented one unrepresented or ah, three represented and none on unrepresented. And notice that the only case here that's missing is where only one of these shifts are represented. And that was actually the answer to part B. So recognized that this is, ah, the compliment to Part B, which means we'll have one minus 0.346 That will give us 0.99 um, 97 in part D. At least one shift unrepresented. So we want one unrepresented to represented or to unrepresented. And one representative, these air The only cases this one is done already. This is our part B answer where only one of the shifts is represented. This one will require a little bit more calculations. We want to try to represent two of the shifts, so that will be calculated as, um, suppose we only want day shift in evening shift. That would be 52 C nine over 67 c nine plus, uh, evening and morning. That would give us 37 c 9/67 c nine plus ah day and morning would be 45 c nine over 67 c nine and we add it to the probability in part B, and when you put all this into a calculator, that will give us the answer of 0.11 zero 0 to 6.

Alright in this problem we're looking at a probability distribution about um the number of days in the summer months that they can't work because of weather. Um a construction group. So Um on part they were finding the probability that no more than 10 days um which is going to be No more than 10 days is the same as less than or equal to 10. And then that's going to be all of the possibilities that are less than Or equal to 10. So we are going to include 10 there, that is still no more, it's not more than 10. So we will include that. Um which if we add up all those possibilities We get 65 Alright. Part be found the probability That from 8 to 12 days. So that's just between eight and 12 Which is going to be 15 plus 20 Plus 19 plus 16 plus 10. Just probably going to be our most likely. Yeah, so about 80%, chance that it's going to be somewhere between um 8 to 12 days lost, for part C. The probability that X is equal to zero. If you look at your probability distribution, that is not an option. And just to check things out here, I'm gonna add up all of our probabilities and make sure That they all total up to one. So there is no option of there being no days loss which we know that's not really going to happen, whether it's always going to play a part, um and then last are mean and standard deviation, I'm going to uh enter that into our calculator, it's a whole lot faster than doing the formulas by hand. So we enter that in do first variable stats And our frequency list is our probabilities. So too, and we get a mean of 9.79 and a standard deviation of 1.86. Now, the interpretation for the mean is that you would expect. It is the expected value of days missed. You would expect to miss about 979 days. So somewhere between nine and 10 days is what you would expect the construction crew to miss over the course of the summer.

Right. It's a little bit confusing to get the events in this problem. But basically we're going to have D. Which is identifying an item as defective and then will have N. Which is the item actually being non conforming. Okay. Now what is the problem? Say we have a 99% chance of correctly identifying defective items. So that's the probability of um identifying an item as defective. Mhm. Now the probability of it being nonconforming given that we identified it as defect. All right. Um I think it's the other way around. It's important to get this correct but we've got to read the wording carefully correctly identifying defective items. So given that it was defective Then we identify it defective with a 99% probability. And given that an item was good. So and prime not nonconforming classifying as defective was point half a percentage 0.5 good. Okay. These probability should not add up to one. This makes sense. Then the other way would have had them add up to one and it would have been wrong. Okay. The company has evidence that his line produces .9% nonconforming items. Um Okay. What does that mean? That means that the probability of N. Is 0.009. Okay. Now what is the probability that an item selected for inspection is classified as defective? So we're looking for the problem probability uh being classified as defective. PFD. Okay. Um Let us look at what this is. This is the same as the probability of D intersection and over the probability event because N. Is the whole world here. And I know the probability of ends. So I can find the probability of D intersection in. So the probability of D intersection N is going to be the probability of D. Given end Times the probability of end zero nine. I'm just gonna use calculator to do this to make sure all the decimals come up correctly. So that is .00891. Okay. Um Same thing here, this is the probability of D intersect and prime Over the probability of end. So probability of d intersect and prime is going to be a .005 Times zero nine Which is .000045. And let's just draw a Venn diagram here we have D. And N in the whole world And the intersection is going to be a .00891. And then the probability of ends with this whole circle Is .009 which means over here we have zero 009. Because now these should add up 2.009 which they do. Okay. And then D. And not N is exactly that region. That's exactly the part that's inside the Circle D but not inside the circle end. So that's .000045. And then remember part A. We were just looking for probability of D. So now we can just add these two together. So what is that zero 8955. or .8955 percent. Almost 1% probability of classifying as defective. Now if a random selected if an item selected at random is classified as non defective. What is the probability that it is good? So what is the probability of not being defective? Um given that we classified it as not effective? Okay so that is um the probability of being not defective. All right probability of being good intersect. Probably have been classified as good over the probability of being classified as good. Okay and we can find all those. So um that number is actually what's on the outside of this Venn diagram. So I just need to take one minus all of those numbers. Which is going to be .990955. Okay so .990955 divided by probability of not being defective is 1- that number. So 0.89551 minus that .991045. And now I'm going to divide the two. Yeah. Which will give me approximately 0.99 99 09 Or that's 99.9909 1%. So the probability of an item being good after being classified as good is pretty high


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