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Question 110 pts1 The mass of a dozen eggs may be recorded individually like this: 5.5.2,5.2,5.2,6,6.1,6.1,8.3,8.3,8.3,8.3,8.3or in 3 frequency chart like this:Mass...

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Question 110 pts1 The mass of a dozen eggs may be recorded individually like this: 5.5.2,5.2,5.2,6,6.1,6.1,8.3,8.3,8.3,8.3,8.3or in 3 frequency chart like this:Mass (grams) Count6.18.3Suppose that one of the 8.3 gram eggs was weighed incorrectly and it should actually have been recorded as 8.5 grams: How will this change effect the median mass?0 (al Thc mcdlan mass did not = change;(b) The median mass changed from 6,1 t0 6.2(c) The median mass changed from 6.1 to 6.(d) The median mass changcd fr

Question 1 10 pts 1 The mass of a dozen eggs may be recorded individually like this: 5.5.2,5.2,5.2,6,6.1,6.1,8.3,8.3,8.3,8.3,8.3 or in 3 frequency chart like this: Mass (grams) Count 6.1 8.3 Suppose that one of the 8.3 gram eggs was weighed incorrectly and it should actually have been recorded as 8.5 grams: How will this change effect the median mass? 0 (al Thc mcdlan mass did not = change; (b) The median mass changed from 6,1 t0 6.2 (c) The median mass changed from 6.1 to 6. (d) The median mass changcd from 6.1 to 7.1 (e) The median mass changed from 6.1 t0 8.1,



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Below is a boxplot of $\mathrm{CO}_{2}$ levels (in grams per kilometer) for a sampling of year 2018 vehicles. Suppose follow-up testing determines that the low outlier should be 10 grams per kilometer less and the two high outliers should each be 5 grams per kilometer greater. What effect, if any, will these changes have on the mean and median $\mathrm{CO}_{2}$ levels? (A) Both the mean and median will be unchanged. (B) The median will be unchanged, but the mean will increase. (C) The median will be unchanged, but the mean will decrease. (D) The mean will be unchanged, but the median will increase. (E) Both the mean and median will change.

So we have a question that is given to us. You can read it. We are just going to get in the numbers away. So the data set has seven values. That is right. Then it is. 55667 11, 11. Okay, so let's try to arrange these because it will help us in finding the media. So the values are five, 5667 11 and 11, Right. 123456 and seven. We have seven different values now. How do we calculate the mean, Mean is going to pay submission of all the observations that I have divided by the number of observations. Right. What is going to be the summation of observations? Five plus 5, 10 plus 6. 16 plus 6. 22. Then we have 29 40 and 51. Right. So this is 51 divided by 123456 and 77 different values. All right, so this is going to be seven point. This will be seven point 28 Great. So let us just say that this is 7.3 well rounded up to one decimal place. So this is I mean, what is the median? Median is the middle value of our data set. So we have seven observations with us. The middle value happens to be this one. The fourth value sixth. Right. In order to find the media, we have to raise that he does it. And either the ascending or descending order. That is what we did. So this over here is six. So our media is going to be six. And what is going to be the mode? The more now out of seven values, I can see that five has occurred twice. Six has occurred twice. 11 has occurred twice, but seven has occurred only once. So in this case, there is a murderer. Multiple modes. So one other modes the values that have occurred two times in this case, so 56 and 11 rate Had there not been a seven, we would say that there is no more right every values occurring six times. Sorry, two times. But right now there is one value which is occurring ones. And there are three values which are occurring twice. So we would say that mode is all the three values that are occurring twice. This is our answer

Well. Technically, the shape of the graph would not change at all. But the verdict costs. Carol will be different, but and the relative height would be the same.

In this video, we've been provided with a set of data values for which we'd like to calculate the median and mode. The first step in this process is to sort the data either from least a greatest or greatest released. So so far I have the data sorted using a spreadsheet, document or program, and this can be done by hand as well. First, let's calculate the median. We'll use the information that there is 30 observations in this data set to tell us that we need to look at observations and divided by two and and divide by two plus one. So observation and divide by two is observation 15 and the very next one is observation. 16. We need to look at both of these observations, since with n equals 30 being even, we must average the two observations with numbers 15 and 16. Let's go back to our sort of data and find those values in our sordid say today. Gah, we have that. The 15th observation is 87.7 no, and the 16th observation is 88 2.0 This tells us that the median is going to be equal to 87.70 plus 88 0.0 Divide by two, which equals 87 point 85 So that's the median for this set of data. Notice that the quantity 87.85 is between these two values, so it splits the data set in half. For the next part of this video, we're going to be calculating the mode of the same set of data. We could use the unsorted data values. But once again, the sort of data values make this process much easier. So the mode is the observation that's repeated the most in this data set. So what we're going to do is carefully go down the line and look for repeats. When we find a repeat will count how many times a quantity is repeated and this will aid us in finding the mode. So it's carefully scanned through this entire data set. And as we scan through, we have a repeat here twice at 85.7, though continuing to scan it looks like Let's see, here we have 88.0 repeated twice, 88.20 is repeated twice next 89 points to zero is repeated twice, so there's still no clear, clear winner of what the moat could be. And if we keep going down, there is no further repeats. Since there's no one value that's repeated the most, there's a four way tie. There is no mode for this set of data.

Question 68 tells us that a carton of eggs is filled with a dozen eggs that I'll have identical mass m. Initially, the center mass is then right in the middle. Um and then a asks, Does the location of the center of masses of the exchange Maur if one egg is removed, or if two eggs are removed, will remember center mass is found by multiplying mass times the position or distance of one thing, plus the mass times the distance of the next thing. Since all these masses are the same, the only thing that really affects this is the distance. Since two is farther away, it does have a greater effect on the distance, then are on the center mass. Then we're supposed to actually find the center of masses of these things. If eggman is removed or a two is removed Party. We're going to part B. Sorry. We're gonna go ahead and remove so we're gonna take egg one out first. Then the center mass and ex measuring from here is given by two m times. Negative 15 right? These eggs are six centimeters apart. Eso were three centimeters of the first take six more is negative. 96 Moore's negative 15 On the other side, we would go three to the first day, and then a six more is nine and then six. Moore's 15 so to M times negative 15 plus two m times Negative nine plus two m times three. Um, plus and this is negative. Three plus m times three plus two m times nine plus two m times 15 You'll notice I don't actually have to do all of this math. The reason I don't have to do all of this math is because this to M minus 15 to m plus 15 to a minus nine to him plus nine. All I end up with actually is this m three is going to cancel one of those negative to injuries. And, of course, there's 11 em down here. So this is really just negative 3 a.m. over 11 m, which is negative 3/11 or if you like, negative three divided by 11 is negative 110.27 centimeters for our X coordinate further why we can do exactly the same thing. Center mass and why, like before all of these will be balanced, the only one that will be balanced is this one. So I'm not gonna keep track of all of these. I'm just gonna go ahead and say we've got negative 1.5 because it's 3.5. This is seven centimeters between them. So negative, 3.5 centimeters times m. And then there's, of course, 11 eggs because we removed ones from negative 3.5, divided by 11 gets us ah, position of negative negative 0.32 centimeters. So our center of mass has been shifted left 0.27 and down 0.32 So somewhere in there and then, um will repeat the process for a two. But again, we're gonna go ahead and just say, Oh, well, we've got symmetry everywhere. The only place we don't have symmetry is for this egg. And so we will have negative 15 times m over 11 him for our X coordinate. And so negative 15 divided by 11 is negative, 1.36 centimeters. And then for why? The only place we don't have symmetry is again right here. And this is going to be the same as that one. We still have negative 3.5 m divided by 11 him, and so this is still negative 110.32 centimeters


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