So we have a set of data that three sets of data and we're gonna be testing each set of data for outliers. So in order to test for outliers, we need to understand the formulas that will will be repaired. So first, we're gonna need to know quartile one. We're gonna need to know court tell three. I'm gonna need to know the i Q r and then the formulas for the upper and lower boundaries. So the lower boundary can be found by taking quartile one and subtracting 1.5 times the i Q. R and the upper boundary can be found by taking Portal three and adding 1.5 times the i Q. R. So our first steps they're going to be to identify our core, tells and are like you are. So if we look at this data, I've already arranged the data in order from smallest to largest, because when finding measures of position, order matters, and so the middle of our data is going to be right here. Since we've got eight data values here, that's gonna be the middle. So to find court, how one I need to find the middle of the lower half of the data. I've got four data values there, so the middle is going to be in the middle of these two numbers. So to find court, tell one, I'm gonna have to take their average So 21 plus 25 divide by two. So 21 plus 25 is 46 then divide by two. And that gives 23. And that's quartile one. And then to find quartile three, I want to find the middle of the top half of the data again, therefore, numbers there. So the middle two numbers are going to the middle of the middle. Two numbers are one to get portal three. So the average of 32 and 34 divide by two others up and divide by two. Or you can just kind of reason through it. What's in the middle of 32 34. And the answer is 33. Okay, so now that I know that I can find the I Q. R. So the I Q R's court L three minus quartile one. So that's 10 and then I need to know this. But 1.5 times the I Q Are so in this case, 10. So 1.5 times 10 is 15. So what confined My lower boundary, my lord Bounder is going to be court Ellen minus 1.5 times 10 So minus 15. So it tells me the lower boundary for my outliers is eight. Any number lower than eight would be an outlier. My upper boundary is going to be my court l three plus 1.5 times the i Q R groups. I didn't need to put just move right there. So plus 15. And that gives us 48. So any number greater than 48 will be considered in that liar. So if we look back at our data, our lowest numbers 19 it is not below eight. So it is not an outlier. And then at our upper end, 36 is our maximum value, which is not greater than 48. So in this data set, there are no outliers. Okay, let's move on to a data set B and again, I've taken these numbers and written them in order from smallest to largest. And first thing we're gonna do is identify our media So we have 123456789 Data values and equals nine. So nine plus one divided by two. That means our median is in our fifth position. So if I look here 12345 This number right here is the medium. That's what splits my data into the lower half in the upper house. So again, I'm gonna identify court, tell one by looking at the middle of the lower, um, data values. So eight right between 82 89 and that gives me 85.5 and then for the court, tell three we're gonna look at the middle of the top half of the data. So in between these two values, so take the average of 97 100 and equal sank. I lost their That's 98.5 and then I confined the i Q r and my boundaries. So, like you are 98.5 minus 85.5 groups. Um, and that is 13 and 1.5 times 13 Yes, 19.5. So, again, I'm going to calculate my lower boundary by taking my core tell 1 85.5 and subtracting 19.5. Which gives me 66 and the upper boundary score. Tell three I'm not to put the value for top three 98.5 plus 19.5. So that gives us 1 18 So then any number lower than 66 would be in that liar, and any number greater than 1 18 would be in that like So when I look at my data values, um 65 is less than 66 So 65 is an outlier, and 101 is my largest value. It is not greater than 118. So 65 is the only outlier in that set, All right, moving on to our last day to set it's a small one. It's a small number of numbers with a really wide spread of numbers. So this one is going to be kind of interesting and also very easy to identify our quartets. So when we look at this, our media, we've got five numbers. This is gonna be your median. And to find our core, tell one it's gonna be the middle of these two numbers on to find the court, tell three it's gonna be the mill of these two numbers. So the average or tell one, is going to be the average of 1 75 and 3 71 and that is to 73. And then for quartz. How three It's going to be the average of 5 27 and 1007 and that IHS 767. So moving on then toe like you are, that's gonna be 7 67 minus 2 73 And that gives 494 and 1.5 times for 94. The I Q. R means that 115 times that number is 741. So we can kind of tell what's gonna happen here already. But we'll we'll finish these calculations. The lower boundary in the upper boundary with 1.5 times the ICU are being such a large number. It would be really weird if we had the outliers here, So, uh, but we'll proceed to 73 minus several 41. Well, that's going to give us a negative number of negative 468 so we know that we don't have any numbers that are lower than negative. 400 sixty's and no low end out liars at the upper boundary. We're gonna take Wartelle three, which was 7 67 and add 7 41 which is going to give us outrageously large number 1508. So, do we have any data values above 1508? And we do not. So not surprisingly, Ah, there are no outliers in this data set.