WEBVTT
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this question gives you a data set and ask you
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to use frequency tables to find that mean the data
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set was given an exercise too. And we made
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frequency tables and exercises two and four. Part a
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wants us to use the frequency table we made in
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exercise too. This frequency table has had been with
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arranges of 10 starting at 30 we had ranges 30
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to 39 40 to 49 50 to 59 60 to
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69 70 to 79 82 89 and 90 to 99
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. And the frequencies we found where 16 13 22
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17 13 and eight. Now we know when we're
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using frequency tables to find a mean we have x
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bar are mean is the sum of X times f
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all over n where n is also can also be
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found to be the son of F So we know
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we have our frequency. That's this column here.
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That's their f column. But what is our ex
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call? What are our values? We can't really
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use this range of values because, Well, how
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would we do that? So, to reconcile this
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, we take X to be the midpoint of each
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of these ranges. I'm gonna write it. Make
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a new column called X with the midpoint of each
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of these ranges, which is 34.9, a 44.9
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and so on 54.9 what we'll do next. Excusing
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that 95 34.5 44.5 from just two races. 5555
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And then we'll have 74.5 84.5 94.5. What we'll
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do next is have another column X times f where
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we just multiply these values times their frequencies to get
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what we need for the numerator. When we do
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this will get a serious numbers. 34.5 in the
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first column, 267 then 708.5 141,419 1266.5, 1098.5
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and 756. Now we have all of our F
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values. Oliver XF values. What we need to
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do now is some them up. I'm gonna make
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it another road down here for the sums. So
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when we sum up all of our frequencies, we're
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going to get 80. That makes sense because we
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had 80 data points to begin with. This is
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what we'd expect, and that's a good check to
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see if you're off in your frequency table. If
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you have a different number of frequencies, when we
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some of our X times F values, we get
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5550. So now all that's left is to divide
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the two. We know X Bar is going to
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be the sum of X f, which is 5007
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or 50 divided by n, which is some of
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F, which is 80 and we'll get that X
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bars equal to 69 0.375 So now let's do the
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same thing with the other frequency table that we that
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we made. Why do we do it twice?
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Well, what's the use here? Um, the
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reason we are checking it like this is because when
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we're using frequency tables to get a mean, it's
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only an approximation. Um, we don't have since
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we're not adding exactly the numbers that we are given
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, Um, we don't have the exact total of
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all the all of our data points. These are
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only estimations of the mean. They require fewer additions
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so that little bit easier. But they're a little
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bit less accurate and can be skewed a little bit
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. Say, if we had different ranges or arranges
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started in different places, things might change. So
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let's take a look at the second, uh,
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frequency table that we made this time. We so
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used ranges of length 10 but they started at 39
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. So we had 39 to 48. 49 to
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58 59 to 68 69 to 78 uh, 79
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to 88 89 to 98 is there are are ranges
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. We also had frequencies. This is from exercise
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for 6 13 20 1913 and nine. Now,
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again, I'm gonna do the same thing. When
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use the mid points for each of these are X
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values will have 43.5 53.5 and so on. And
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then I multiply each by the frequency so we'll have
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our x f column. Six times 43.5 is,
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uh, 261 to 61 will get 695.5 for our
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next one. Then 1270 then 13. 96.5,
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1085.5 and 841.5. So now again, well,
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some of the f sent all the X efs All
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the F's will some to the same thing. We
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still have 80 data points, so and is 80
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and then when we some are accepts this time we
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also get 5550. Now, this is just by
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coincidence that it was possible to get a different some
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of X efs. But in this case, we
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got the same thing when we took two different frequency
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tables. So now we need to do is divide
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. We know export is gonna be 5550 over 80
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which again is 69.375 And there's your fine