Hi, I'm Dr. Shah, I was the National Lecture Competition winner in 1989, and I am the Maths master at Mathscool. Now ready for a new word in maths? Now, there are three different cases in which we need to calculate 'median'. The first case (1) is that we have a list of data, the second case, (2) is that we have a table of ungrouped data, so that it would probably be a discreet data, something like the number of times you enter the cinema, or number of seats in a car, things like that which you can pinpoint like one, two or something like that. And then (3), a table of grouped data and the method we use would be different for each of these. As soon as we got the marks of class achievements, and say 4 of the students get between 2 – 5 marks, 18 of the students go between 5 and 10 marks and so on. So these types have been grouped, so if we take these 4 students who have got 2 – 5 marks, we don't know any more what they got, may be one of them got 2 marks, and another got 5 marks, may be another got 3 marks, we won't know, because the data has been grouped. So we can't actually find the exact median. All we can do is now estimate median. The way we do that therefore, is we have to change this to a cumulative frequency table. So we change that into a cumulative frequency table, on the top, a slightly different this time, is the upper class boundary

. So it is the top number of each of these classes is the upper class boundary, so we have 5, 10, 15, 20, 25 and 30. Another one in the bottom is against capital F, which means the cumulative frequencies, so the first one is just 4, the next one is those two added together, say 22, all these added together say 44,all of these added together, so 71, all of these added, 90, and then the last one makes a 100. So in this case N=100, that means 100 pupils there altogether in that data. And we want to find the median, is half the brackets N +1, 1/2(N+1), which is going to be a half a hundred plus one by two, which is going be a fifty and a half,[50 and 1/2] value. Now this time because N is large, for that we find 50 and a 1/2 value, it can be approximated into a value of 50, I don't think anyone would really mind that mark, say when you have a large number, for example, bigger than 30, then you can approximate it, instead of having a 50 and a 1/2 value, approximating it into a 50 value, that would be acceptable. To find this value, we don't use the same method as we used for our table of ungrouped data. We don't just look for 50 in the table and then read out, we actually find what the number is in between these. So to do that, we have to draw a cumulative frequency curve. So to draw a cumulative frequency curve, we can have the upper class boundary here and frequencies here, so the first upper class boundary is 5, and the frequency would be 4, next one is 10 and 22, so 10 and 22, and so on. In that way, we can connect those points to form a curve. And I have done one for this table already here, to show that what it would look like. So I plotted those points. I connected them together to form a smooth curve, and the last point is when 'x' is 30, which is that point there, from there the curve goes flat, curve goes flatter as it reaches the top. And you can see over here it is written N is 100 where the curve connects it. And I wanted to find the 50th value, so the 50th value would be here, at 50. Now what I would need is to do connect the line across back here and down to work on to see where the 50th value is, and again I have done it on a separate graph, and you can see from my graph, going 50 over here, across the curve and down here gives me, 16. So the median for this data is 16. .

. So it is the top number of each of these classes is the upper class boundary, so we have 5, 10, 15, 20, 25 and 30. Another one in the bottom is against capital F, which means the cumulative frequencies, so the first one is just 4, the next one is those two added together, say 22, all these added together say 44,all of these added together, so 71, all of these added, 90, and then the last one makes a 100. So in this case N=100, that means 100 pupils there altogether in that data. And we want to find the median, is half the brackets N +1, 1/2(N+1), which is going to be a half a hundred plus one by two, which is going be a fifty and a half,[50 and 1/2] value. Now this time because N is large, for that we find 50 and a 1/2 value, it can be approximated into a value of 50, I don't think anyone would really mind that mark, say when you have a large number, for example, bigger than 30, then you can approximate it, instead of having a 50 and a 1/2 value, approximating it into a 50 value, that would be acceptable. To find this value, we don't use the same method as we used for our table of ungrouped data. We don't just look for 50 in the table and then read out, we actually find what the number is in between these. So to do that, we have to draw a cumulative frequency curve. So to draw a cumulative frequency curve, we can have the upper class boundary here and frequencies here, so the first upper class boundary is 5, and the frequency would be 4, next one is 10 and 22, so 10 and 22, and so on. In that way, we can connect those points to form a curve. And I have done one for this table already here, to show that what it would look like. So I plotted those points. I connected them together to form a smooth curve, and the last point is when 'x' is 30, which is that point there, from there the curve goes flat, curve goes flatter as it reaches the top. And you can see over here it is written N is 100 where the curve connects it. And I wanted to find the 50th value, so the 50th value would be here, at 50. Now what I would need is to do connect the line across back here and down to work on to see where the 50th value is, and again I have done it on a separate graph, and you can see from my graph, going 50 over here, across the curve and down here gives me, 16. So the median for this data is 16. .

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