Good morning. Uh, this question that we have in front of us right now offers us the enrollment for 35 community colleges across the United States and asks us to do several things. The first thing that it asks us to do is to divide the data up into five even intervals. Um, call that classes in statistics. And so, in order to do that, the first thing we need to do is identify the maximum value and the minimum value. The maximum value for this data set is 28 1001 165 and the minimum value is 1263. By subtracting the mom from the maximum, we find our range, the range of our data. How Why does it spread? And the spread his 26,902 to divide that into five. Even inter rules we divide by five, giving us 5380.4 Now. Um, what we're doing when we divide by five is we're trying to determine our class with health. How big are our classes? And so we always want to take this number even if it was 5380 even. And round up to the next number, we round up 5000 381. That gives us our class That told us what we're gonna add to our first class limit to get to our next one. So for five intervals, we start with our minimum data value. Okay? And we take that 5381 to determine what the beginning of my next class is next, even if get 6644. So I add 5381 again for 12,000 in 25 and added again for 17,000 406. And my last class is going to start with 22,000 787. Um, those air the lower limits for each one of my classes came, um, In order to determine the upper limits for each one of my classes, I start one below the 2nd 1 So you're toxicity for just a minute? Um, one below that is gonna put me at 6643. Okay, One below that. 12,000 and 24 17,405 and 22,000 786 Now for my upper class limit. I just take this class with here, um, and I add it to this one right here and get my last one for 28,000 167. Oh, now, this is where my five even hold 12,060. 343,644 to 12,000 and 24 so on now, too organized. The data into those five categories are even intervals. We need to basically tally up how many pieces of data fall into each category. Um, and so the first piece of data that we see it says 6400 in 14. Well, that would fall here in this first class, and then the next piece of data is 50 1550. That would also fall in that first class. And so we go through and basically tally up. Um, how many pieces of data falling to each class We do that correctly. We should figure out that 19 pieces of data pull into that first class nine going to the second one in the third. We're in the fourth, two in the fifth, and we can double check our work by adding up all of these all of these pieces of data and well, if we get 35 then we did something correctly. Hopefully, right. Um, because there were 35 pieces of data given to us initially. So now part B of this question asked us to construct a history, Graham and, um, to construct a history. Graham, With this data, there's couple different ways that people do it. But one thing that's common is the frequencies always go on vertical act frequency. So in this case, that would be how many colleges? Ah, um, intervals are enrollments within each one of those intervals. And so here I go, to see the highest number of frequencies that I have is 19. So I think I'm going to go by fives. And when you set up this vertical axis, it's really important that won her lowest numbers at the bottoms into so an equal increments. So if I'm going to go by fives, I need to go five and 15. I want you like that. I can't. Um change the intervals between tally marks as we go up. And now along the bottom of my history. Graham, some textbooks will call for cross foundries. I'ma call for midpoint. I'm in this case. I think we're gonna do class boundaries. Um, so class boundaries basically require us to meet halfway between classes so that they can share a bar because one of the our share a limit here a boundary? Um, because one of the requirements of the hist a gram is that when we construct our bar graph, which is what instagram is, the bars touch. And so I would have been a do is with my lowest class limit, which was 1263. I am going to subtract five, 1000 262.5. And now, with the upper class limit, I'm going to add point by 66 43. Wait, Fry. And I'm gonna do that with all of my upper class limits on down the line here. So I'll have 12,000 24 white farm, and then I will have 17,000 of 18,000 400 and 5.5 22,000 786.5. And last but not least, I have my for most class limit. 28,000 167. Okay. And so now the bar that we construct right here is gonna represent the dead between those two enrollments. And we had 19 colleges, it had, um, enrollments between 1263 and 6000 643. So we create a bar that goes not quite up to 45. The little cricket do better on paper, I'm sure with a ruler if you need it. But, um, my next class has nine pieces of data in it, and so I'm gonna go just below 10. Men draw bar, that's a little better. And then, um, I have just one enrollment between that 12,000 and 24 12,000 and 25 then the 17,405. I have Forbes mongo little bit higher, but not quite to five that are. And then too for my last are your Houston. Graham should look similar to that. Um, the next thing it asks us is if we were to build a community college, which would be more valuable. The motor that mean you go through this data, there's not one piece of data that appears more often to the other. But when we find the mean, I think we'll find that it's not necessarily representative, uh, most colleges either. So when we look at the history, you can tell this class is the one that has the most occurrence. And, um, that would be more of most community colleges. In order to find the mean, we want to use either a graphing calculator or one that allows you to enter all of our enrollments into it into a table and then find our one variable. So once you do that, um, you can look at your one variable statistics and you'll notice our ex bar or as we know in statistics, are our represents our sampling that is 8628.7. Other piece of information. Look for here. Let her e. S the strip you is our sample standard deviation. So when we're looking through our one variable statistics, we want to look for the letter s, um because that represents a sample standard deviation bond. That would be in this 6000 943 believed eight s. So we now have our sample mean in our sample standard deviation on the next part that it asks us for is how many standard deviations with the data value or the enrollment of 8000 be away from the means. How far away is 8000 from the mean in terms of standard deviations? And so, with statistics, we call back a Z score. Okay, Ours thescore tells of how many standard deviations away from the mean of specific data value is, and there's a formula for that. Um, it looks like this where we take our data value X. In this case, that's 8000. We subtract the mean, which in this case is 8628.7. And then we divide by our standard deviation 6943. I want A That gives us a Z score of negative 0.9 What that means is that, um, first off, a negative Z score tells us that the data value we're evaluating is below demean. And since it's negative 0.9 It tells us it's not very far below the mean. It's not even one standard deviation below the mean. It's just a little bit below me. And you can see that if you look at those numbers right, we're evaluating 8000 and, um, our means 1000 628. Our standard deviation, Uh, 6900 right? That's 600 difference. There between, um, the 8000 and 8600 is not much compared to that 6900. Almost seven standard deviation. That should complete this question.