Modal Class of a Histogram with Unequal Class WidthsDate: 07/12/2007 at 18:13:11 From: David Subject: Modal Class of a Histogram with unequal class widths Why is the modal class of a histogram with unequal class widths the bar with the highest frequency density? This gives a different answer when you look for the modal class from a frequency table. I am confused because I have always been taught that the modal class is the one with the highest frequency from a histogram or frequency table. Date: 07/12/2007 at 23:51:02 From: Doctor Peterson Subject: Re: Modal Class of a Histogram with unequal class widths Hi, David. Many treatments of histograms assume equal class widths; that is probably why you were taught what you were. When classes are allowed to vary in size, we have to make some adjustments to the whole concept of a histogram. When classes have unequal widths, the vertical axis of a histogram must represent not frequency (number of occurrences) but frequency density (frequency divided by class width), and the class widths must be accurately represented on the horizontal axis, so that the AREA of each bar (not the height) represents the frequency of that class. Therefore, the modal class is indeed the highest bar; but that height is not the frequency you read from the frequency table. To see why it makes sense to use frequency density in a histogram, consider this example: Score Frequency 0 - 50 25 50 - 60 10 60 - 100 20 If we make the histogram based on frequency, and ignoring class widths, it looks like this: 25+-------+ | | 20+ | +-------+ | | | | 15+ | | | | | | | 10+ +-------+ | | | | | 5+ | | | | | | | +-------+-------+-------+ 0-50 50-60 60-100 It looks like there are a lot more low grades, which is true. But WHY are there so many more of those? Because the low grades form a larger class! Perhaps if we had equal intervals it would look like this: Score Frequency 0 - 10 2 10 - 20 3 20 - 30 5 30 - 40 7 40 - 50 8 50 - 60 10 60 - 70 9 70 - 80 6 80 - 90 3 90 - 100 2 25+ | 20+ | 15+ | 10+ +-----+ | +-----+ +-----+ 5+ +-----+-----+ | | +-----+ | +-----+ | | | | | +-----+ +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 That gives a much different appearance--now the mode is clearly at 50-60, and all because we made a different choice of intervals. We'd like to make our original histogram look as much like this as possible, to make it independent (to some extent) of the inequality of classes. Since we don't know all the numbers for the equal classes, the best we can do is to take the total of the classes from 0 to 50, and from 60 to 100, and spread them out evenly--which amounts to averaging them: Score Frequency Average 0 - 10 2 10 - 20 3 20 - 30 5 30 - 40 7 40 - 50 8 25/5 = 5 50 - 60 10 10 60 - 70 9 70 - 80 6 80 - 90 3 90 - 100 2 20/4 = 5 25+ | 20+ | 15+ | 10+ +-----+ | | | 5+-----+-----+-----+-----+-----+ +-----+-----+-----+-----+ | | | | | | | | | | | +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Averaging the frequencies in the little intervals is what the frequency density means, except that we will divide not by the NUMBER of equal intervals, but by the SIZE of the intervals (which is proportional to that). Our corrected histogram looks like this: Score Frequency Density 0 - 50 25 25/50 = 0.5 50 - 60 10 10/10 = 1.0 60 - 100 20 20/40 = 0.5 2.5+ | 2.0+ | 1.5+ | 1.0+ +-----+ | | | .5+-----------------------------+ +-----------------------+ | | | | +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ 0-50 50-60 60-100 Now the height of each bar TIMES its width (that is, its area) gives the frequency, so that the visual impression is accurate; but the highest bar (the modal class) shows where the data is most concentrated, rather than just which class we happen to have chosen to take in most of the data. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 07/13/2007 at 05:05:26 From: David Subject: Modal Class of a Histogram with unequal class widths So, if you use a frequency table that has unequal class widths to work out the mode, you could get a different modal class to the one found by a histogram. So with unequal class widths we could get different modal class depending on which method is used. Date: 07/13/2007 at 10:54:41 From: Doctor Peterson Subject: Re: Modal Class of a Histogram with unequal class widths Hi, David. If you defined modal class as simply the class with the greatest frequency from the table, then you can get different results than when you use the proper definition, which, like the histogram, is based on frequency density rather than frequency itself. This is why we do not define modal class that way. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 07/13/2007 at 12:57:43 From: David Subject: Thank you (Modal Class of a Histogram with unequal class widths) Thanks for your help. I now understand that the definition for a modal class either from a Histogram or Frequency Table, is the one which gives the greatest frequency density. But why is the modal class the one with the highest frequency density? I can see it for equal widths, but do not quite see it for unequal class widths. Regards David Date: 07/13/2007 at 23:07:51 From: Doctor Peterson Subject: Re: Modal Class of a Histogram with unequal class widths Hi, David. You seem to be asking the same question you started with, so perhaps what I said was not convincing with regard to the modal class. There is a reason for that: this is really a matter of definition, and there are two reasonable ways one could define "modal class" as an extension of the idea of mode. First, we can just take the modal class as the one with the greatest frequency, and not worry about the ideas I showed you. Then the modal class will not be the one with the highest bar, but the one whose bar has the greatest area. Second, we can take the histogram as the key idea, and take the modal class as the one with the highest bar. My discussion of why we define a histogram as we do is the justification for this. I have seen BOTH definitions used! Presumably, from the wording of your question, you have been taught the second definition, which is the one I prefer. If you don't like it, then it's probably because you prefer the reasoning behind the first definition, which is to keep things simple, modeled after the ungrouped mode. You'll just have to recognize that your text is using the other approach, and take their definition as given. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 07/14/2007 at 12:30:44 From: David Subject: Thank you (Modal Class of a Histogram with unequal class widths) Thanks for your help. I now understand and will be using the frequency density definition. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/