Remember that, the way to find the median of a data set, is to put all of your numbers in order from smallest to largest, and then find the middle value. If there’s two middle values, because you have an even number of values in your list, take the average of then middle two.
We are going to see that here. Find the median of this list of numbers. My first step, always when finding the median, is to put them in order from smallest to biggest. Sometimes it’s done for you, sometimes it's not. That’s a drag.
Before you guys do this homework, if you are really good at excel, if you can type these numbers into excel. Excel will put them in order for you. Also a graphing calculator, you could type them put them, it will put them in order for you using matrix.
I’m going to start writing them in order. I’m looking through and I see 12, and then 25 is bigger than that. So I’m going to leave some space to write 25 out here. A 9, 60 is way the heck out here 15 okay 20 is about there-ish and 18. Not perfect but at least I have them in order of increasing value. Now I’m going to move that 60 back over. So the way to find the median, once you have them in order from smallest to largest, is to find the middle value.
But if I try to find the middle here by going like from outside in. I have two middle values you guys see that? I have like 3 on this side, 3 on that side. Those two are stuck in the middle because I have an even number of numbers. 1 2 3 4 5 6 7 8, I have 8 values that means there’s not going to be only one that’s stuck at the middle.
So the way to find the median, when you have an even number of numbers, is to find the average of those two or the mean. 15 plus 18 divided by 2 will give me my answer. So let me grab a calculator as you got here. 15 plus 18 is 33, divided by 2 and you get 16.5.
That’s the median or the middle of my data set. I think in this problem it's kind of interesting to think about. Because, if I were to do the average of these numbers, this high 60 will mess my average way up. It will pull my average really high.
The middle described by the median however, is more accurate. 16.5 is in the middle of this list. Whatever my average would be, will probably be somewhere up above 20 pulled toward the 60. Because 60 is what we call an outlier. It makes this dataset skewed toward this high value.
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