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Question

The boxplo o n Delot {09u roM Ine heignimeles siedWhai = UnexnumdenInalcORC & Hnu103 ?Tho nunmum heigt I6 ha ahl iIhe firs| quartile 0,Iho socond quor 0 (or tho modtan) %Ihe Urd quartilecMAandehe MuIMum

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Identify the minimum, first quartile, median, third quartile, and maximum of each data set. Then make a box-and-whisker plot of each data set. daily attendance: 29 24 28 32 30 31 26 33

This. Russia asked us to make a box and whisker pot. What we know is that we have the least value in the data status. 48 we have the first court child. The median of the lower half between 48. 54 84 is 54 then we know the median is 103rd quarter mile. The median of the upper half is 1 88 goods between 1 24 and 2 56 and the greatest values obviously to 56.

And this question we are given that men of the upper quartile and lower quartile is given to be 10. Okay, Cuban is the lower quite i. and Q three is the upper part I. So let's say this is our first equation. Okay, okay, now simplifying questionable. First it comes out to the Cuban plus Q three equals to 20. Okay, now it is also given that semi inter quartile range. Okay, which is given by Q three minus Cuban, divided by two. This is given by five. Okay, it is given to be five. So just simplifying it to three minus Cubans. Come out to Britain. Okay now we'll add equation number one and the question number two. This was a question number one. So we'll add these two equations. I just write it here. Okay 2 3 is equal to 20. Okay, this was our question number one. Now we just add these two equations to times of 23 comes out to be happy. Okay, which implies that the value of 2, 3 that is upper quarter. He comes out to the 15 and putting developed 23 in question number one, or equation number two, we get Cuban equals 25 Okay, so the lower quarterly comes out to be five and a quarter comes out 15. So option number D is the correct option.

In this problem we are trying to construct a modified black box plot which will account for any outliers in the data. So before we can drive box plot we will have to rearrange our data in order from smallest to largest. So we would start with two seven eight nine nine, 10, 10, 11, 11, 12, 12, 13, 15, 16 and 24. Yeah. So we happen to have 15 items. And in order to do a box plot we will need to find you're three chortled. So we need quartile one, Quartile two and Quartile three. And we start by finding quartile to which is the median of the data. So we are trying to find the middle number in our data. So since there's 15 numbers, the middle number would be the 8th number. So this would be cute to because it's separating the data into eight numbers or sorry, seven numbers lower than it And seven numbers higher than it. So we've got Q two Now, Q one is going to be the median of the first half of the data and Q three is going to be the median of the second half of the data. So the first half of the data happens to have seven numbers. So the middle number would be that number nine Again leaving three before it and three after it. and Q three will be 13 because we now have three numbers before it and three numbers after it. So now that we have found our three chortled, we want to look for outliers and to determine outliers we can use little formulas and an outlier is going to be a number either at the upper end or the lower end of the data that just doesn't fit with the rest of the data. So we can find outliers at the low end of the data By using Q 1 -1.5 times the inter quartile range or the i. q. r. So let's talk about finding the like you are, the I Q. R. Is found by taking Q three minus Q one. So in our example our IQ. R would be 13 -9 or a value of four. So if we want to find outliers on the low end of our data we would take Q. Three or sorry Q one which in this case was nine minus 1.5 times the I. Q. R. Which is four. So we would end up doing nine six or three. So any number or data less than three? Yeah is an outlier. So therefore The number two which is less than three is an outlier. Now to find outliers on the upper end we can use Q three plus 1.5 times the Q. R. So Q three in our example was 13 Plus 15 times four Which would be 13 plus six or 19. So any data greater than 19 is classified as an outlier. So 24 is greater than 19. So this is also an outlier. So now we're ready to create our modified box plot in order to do so we are going to draw a number line and the number line has to account for every number in our data set. So Our lowest number in our data set is a. two. And we've got to get all the way up to 24. So we'll have (345 678 910 11 12 13 14 15 16 17 18 1920 21 22 23 24. Now in a typical box and whisker plot we would plot the minimum and the maximum into the box plot. But in a modified box plot which demonstrates or shows you any outliers, we're going to use asterix to represent are outliers. So we're going to put an asterix at two And we're going to put an Asterix at 24. We're then going to plot our three core titles. So Q one was a nine. Q two was an 11. Thank you. Three was a 13. We're going to create the box centered around those court als and traditionally the Whiskers for the box and whisker plot would extend all the way to the minimum value and the maximum value. But because our minimum and maximum values are outliers, we're going to expend extend our whiskers to the next numbers on those ends that are not classified as outliers. So our whisker would extend back to seven on the low end and up to 16 on the high end. So that is our modified box plot for this data.

In this video we will be making a box and whisker plot, finding the inter quartile range and then answering a question in regards to our information. So I have put the numbers here in front of us. There are 50 items in this particular situation. So that means My median will actually, let's start with us. Let's start with our low value. They're already in order from least to greatest. So my lowest value part of my five number summary will be five and my highest value will be 15. I also need Q. one, the median and Q two. So because there are 50 items that means the median will be 25 units from either side. So 123 And then if I count 25 back from 15, I'll also end up here at 10. So that makes it pretty easy to find the median because even though it's I would normally average the two because they're both 10 just means the median is 10. So now I need to find the middle of the bottom quarter or the lower quartile. So now I've got 25 items and 25 divided by two is 12.5. So that means I'm going to count up 12 so one And then I'm going to circle the next one because that's gonna be my Q1. And just to make sure 34. Okay So nine will be my Q. one, my lower quartile. And then I'm going to do the same thing on the upper quartile and then just make sure sure enough yep there's the 12. So on my number line I see my low values is five. My high value is 15. So I could just start my number line at five 10 15 so I could do that. So I'm going to put a. five. A data about 15. Then Q nine or Q one is 9. So that would be here 11 12. My meeting is 10. Make a box whiskers and draw my bar to represent the medium. Now I need to find the inter quartile range and remember that's the difference between these two or the length of my box. So the inter quartile range will be three. And then finally I am told that Wyoming has a dropout rate That's 7% or around 7%. And I want to know where that is on my graph. So I look at my graph and I see 7% would be right here. So that would fall in what we would call the first quartile because it's below Q one so 7% is in the first quartile.


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