5

98 89 96 96 83 82 34 38 45 36 55 91 36 66 51 98 97 86 43 8854 08 56 56 8 N = 94 87slulzw 17 1820 21 22 23 24 25 26 27 28 62 30 31 32 3362 68 46 67 65 45 6 N & T...

Question

98 89 96 96 83 82 34 38 45 36 55 91 36 66 51 98 97 86 43 8854 08 56 56 8 N = 94 87slulzw 17 1820 21 22 23 24 25 26 27 28 62 30 31 32 3362 68 46 67 65 45 6 N & Test_ScoresType here to search

98 89 96 96 83 82 34 38 45 36 55 91 36 66 51 98 97 86 43 88 54 08 56 56 8 N = 94 87 slulzw 17 18 20 21 22 23 24 25 26 27 28 62 30 31 32 33 62 68 46 67 65 45 6 N & Test_Scores Type here to search



Answers

$$\begin{array}{lrrrr} \hline 13.3 & 9.0 & 11.1 & 9.1 & 8.4 \\ 15.6 & 8.1 & 8.3 & 13.0 & 17.1 \\ 16.3 & 13.5 & 8.0 & 15.1 & 5.8 \\ \hline \end{array}$$

And this question we're gonna be testing to see if there are any outliers in this set of data. So the way we do that is by first identifying what quartile one and Court L three are and then finding the i Q. Are multiplying the i Q or by lump 30.5 and adding that and subtracting that to find our upper and lower boundaries. So let's take a look at this example or this problem. So first thing we need to be able to do here is to identify our sample size intermedia. And so we have 1234567 numbers. So that means that this number right here in the middle 86 is going to be our medium on our medium is important to help us identify where our court tells our so court tell one then is going to be the number that is at the middle of the lower half of the data, and that's easy to see them. That quartile one is gonna equal 84 our numbers air just point out our numbers are in numerical order here. Before we do any of this, and then we look again at the top set of numbers here. And we see that right in the middle of the top set of numbers is our portal three, which is 97. Now, when I look at these numbers, they don't seem to be all that spread out. So I would be really surprised if there were outliers here. Thes number seemed to be fairly evenly spaced out. Nothing to, um, outrageous. But we're gonna test anyway. So next thing I need to calculate is the i Q r and the ICU are is the difference between court l three and quartile one large number minus smaller number. So 97 minus 84 we end up with and I q r of 13 now are out. Liar Test says that anything that's more than 1.5 more than 1.5 times like you are above court l three or 1.5 times I Q are below quartile. One is going to be an outlier, so we need to know what 1.5 or 1.5 times 13. It's how far away from either of those court tells is going to make something unusually high or low, so 1.5 times 13 is 19.5. So our upper boundary and I'm gonna call that you be upper boundary is going to be our third quartile, plus 1.5 times the i Q. R. So in this case, that's going to be 97 plus 19.5, and that equals 116.5. So anything above 116.5 would be considered an out liar. We also have the lower boundary. What would be an outlier at the low end on? That's gonna be anything that is below quartile won by more than 1.5 times the i Q. R. So in this case, that would be 84 minus 19.5, which equals 64.5. So if we look at this data set, do we have any numbers above 1 16.5? The answer's no. Do we have any data values below 64.5? And the answer is no for this data set. Then we would say that there are no out liars. All right, so let's move on to data set B, which I have also put in numerical order and data said be on Lee has six data values. So when we're looking at these six data values, the median is gonna be smack dab here in the middle. It doesn't really matter what the median is gonna be. Be halfway between those two numbers, but it splits our data set into the upper half of the lower half, which makes it very easy for us to identify what our core tell one and our court l three are so cartel won a court l three are those numbers, and we can quickly identify then are like you are, which is gonna be then Q 31 25 minus Q. One at 1 18 And that gives us our I cure of seven. And then we can calculate 1.5 times seven 1.5 times seven is 10.5. So if we go above court, tell three by 10.5 or below court, tell one by temple if I will run into an outlier. So our upper boundary is 1 25 plus 10.5, which is 1 35.5 And if we look at our data set 1 45 is above 1 35.5 So this data value right here is an outlier. That just means it's unusually large compared to the rest of the days. Are lower boundary is our core. Tell one minus 10.5 and that gives us one or 7.5. And if we look, there are no data values below one or 7.5, so we don't have any outliers at the lower end. So there is one outlier in this data said, and it is at 1 45 And then our last data said that we're gonna look at is data set C. We have a few more numbers here on this time we have. How long? 23456789 numbers. Inter data set. Our sample size is nine. So the middle is going to be the fifth number in the sense, so that right there is our median. And so to find our portal one, we're gonna look at the middle of these four numbers here. So we have two numbers in the middle 14 and 15. So to find out what court tell one is we have to take their average, um, so you can take their average and divide by two. But when the numbers are this close, 14 and 15 it's really just the number that's halfway between them. So conceptually you can think through that that it's going to be 14.5 and then at the top end of the numbers you want that middle, the number that's in the middle. So the number that's halfway between 20 and 27. So the average of those two numbers eso 20 and 27 is 47 when we divide that by two, we get that quartile three is equal to 23.5, so now we can move on to calculating are like you are quartile. Three. 23.5 minus 14.5 gives us our I Q. R of nine and then 1.5 times nine is equal to 13.5 and then following the same on process we did earlier. We can calculate her upper boundary and are lower boundary so our upper boundary is going to be our core. Tell three. 23.5 plus 13.5, which gives us 40 cents. All right, 37 and then our lower boundary is gonna be 14.5, minus 13.5, which is equal to one. So if we look back at our data set, do we have any numbers above 37 or any numbers below one? And we can see that In fact, we don't. So in this data set, there are no borders.

In the given question We have a table in which values of N&A 4th an argument and And a 4th n. The values of N are given us 1, 2, 3, 4 and five. And for each value of N, we have the corresponding value of a four pin Which will be our sequence. And this problem. So 149 16 and 25. So let's grow table around this. So this is the table that is given in the christian. So the sequence is given by each door of a 4th and right. So in N equal to one, we will have 1/4 1 which will be the first time, that is one. The second term will be for the third term will be nine and so on. So the sequence that we had Taking into consideration as 14 9 1625. And to determine whether we are told to determine whether the given sequences, arithmetic or not. And to do that. What we do is to find the difference between the constitutive terms in the sequence and if the difference between them are the same, then we can call it and arithmetic sequence. So let's take the first, the 2nd and 1st term right four minus one gives us a difference of three. Next up we have 9, 9, 4, the next constitutive pair. So we have nine minus food, which gives us A difference of five and next we have 16 and 19. So we can see over here that each constitutive term is different by a different number, Right? -9 is seven. Right? So there is no common difference in being between the constituted terms of the sequence. And since there is no common difference, we can say that the given sequence is not arithmetic. So if it were arithmetic, then each of these differences. That is the difference between each consecutive term would have been the same. But over here it's not common. It's different. So since there is no common difference, we can say this is not and arithmetic sequence. So this is how we determine whether a given sequences, arithmetic or not. I hope you understood the method. Thank you. Uh.

Okay, We've got to do this product here and we're gonna do it by multiple and 8.3 times every single one of these numbers individually, 8.3 times negative one is negative. 8.3 a 0.3 times 0.3 is 2.49 a point. Three times 6.2 51.46 a 0.3 times a point, too Is 68.6 going 06 and a 60.3 times a negative. 4.1 is a negative 34 0.3 and a 0.3 times 9.5 is 78 0.85.

We wish to find a formula for this sequence here. So we can actually look at our dominator and our new murder separately. So if we look at our enumerators, that's 12345 which is actually just our end. Right? Because and this is the first term, second term, third term, fourth term. That's true. And they match up exactly with these enumerators. Now our denominators are just um all the odd industries. So which means this is going to be two Times one Plus 1. This is two times two plus one, this is to titus three plus one, this is two times four plus one and the last was two times five plus one. So therefore our formula is going to be equal to our numerator is just an and our denominator is two times and plus one.


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