Modal Class of a Histogram with Unequal Class Widths

Date: 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.