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Class Limits, Midpoints, Frequencies

  1. Jan 27, 2013 #1
    I have a list:

    85 45 75 60 90 90 115 30 55 58 78 120 80 65 65 140 65 50 30 125 75 137 80 120 15 45 70 65 50 45 95 70 70 28 40 125 105 75 80 70 90 68 73 75 55 70 95 65 200 75 15 90 46 33 100 65 60 55 85 50 10 68 99 145 45 75 45 95 85 65 65 52 82

    Sorry for the poor formatting, but I created a program that would count the frequencies, etc. and I still am getting the problem wrong.

    The class width should be 28 since the problem specified that it needed to be divided into 7 classes.

    The class limits make sense to me since the smallest value is 10 and you just need to add 28 (the class width), and the next should be 48, right? Or am I completely off base?

    Class boundaries will make sense once I get the right values for the class limits.

    The midpoint should be the lower limit + the upper, so wouldn't the midpoint of (10 + 38)/2 be 24??

    And relative frequency is just the frequency divided by the total frequencies, right?

    So might you know why the class limits aren't 48-76, etc.? I'm trying to figure out how to work this frequency table. Thanks for your help :/

    http://www.flickr.com/photos/92711644@N08/8421983384/ (sorry, image linking didn't work for me)
     
    Last edited: Jan 27, 2013
  2. jcsd
  3. Jan 27, 2013 #2

    haruspex

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    Where do you get the 28 from? I feel I'm not seeing the whole question.
     
  4. Jan 27, 2013 #3
    Oh, I'm sorry, I forgot to mention that the problem specifies to split it into 7 classes.
     
  5. Jan 28, 2013 #4

    Mark44

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    It would be helpful if you gave us the information you were given or have learned, such as how many numbers are in the list? If there are N numbers in the list, and you group them into 7 classes, each class will have about N/7 numbers in it.

    Sorting the list would be helpful. From the sorted list you can look at each class and see the largest, smallest, and can calculated the class boundaries and midpoint of the class.
     
  6. Jan 28, 2013 #5
    Hmm, okay, so it looks like there are 73 numbers in the list. I found the class width by subtracting the smallest data value from the largest and dividing by 7 (200-10)/7, so around 28.

    From what I can tell, it looks like 10 is the smallest data value in the list, so I would first add 28 to 10 for the first class limit to be 10-38, then add 10 to 38 and add 28 for the next class limit to be 48-76. But is that wrong from then on?

    The midpoint should be found by adding the lower class limit and upper class limit and divide by 2. Wouldn't (10+38)/2 = 24?

    Thanks for all of your help so far.
     
  7. Jan 28, 2013 #6

    Mark44

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    I don't think this is right. You need to put your list in sorted order, from smallest to largest. Since there are 7 classes and 73 numbers, each class will have about 10 numbers in it. 73/7 doesn't come out to an integer, so I'm not sure how you're supposed to deal with this. One way would be to put 11 numbers into each of the first three (the smallest three) classes, and put 10 numbers into all the rest. Your book might have some examples that show what to do in this case.

    The class limits will be the smallest and largest numbers in each class. The class boundaries will be numbers that aren't in your list, and will be the number halfway between the upper limit for a class and the lower limit for the next larger class. For example, if the upper limit for class 2 is 38 and the lower limit for class 3 is 42, the boundary will be (38 + 42)/2 = 40. If the upper limit for class 2 is 38 and the lower limit for class 3 is 43, the boundary will be (38 + 43)/2 = 40.5.
    That sounds reasonable.
     
  8. Jan 28, 2013 #7

    Ray Vickson

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    There is no reason to require roughly-equal occupancy of each class. For example, frequency histograms often have significantly different occupancies in different classes. Are the classes in this problem supposed to have the same widths? The OP did not say.
     
  9. Jan 28, 2013 #8
    Hmm, well, I would guess that they would need to be the same size? This is a problem from chapter 2 of the book, so it shouldn't be too complicated. Webassign said that the class width is correct (28), and that 10-38 for the first class is the right set for limits.
     
  10. Jan 28, 2013 #9

    Mark44

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    What numbers are in the next class? They should be the next larger group of 10 or 11 numbers (assuming you have sorted your list). The class limits will be the smallest and largest numbers in that class. You don't calculate these, as you seem to think - you get them by observation.
     
  11. Jan 28, 2013 #10
    Here's the (assumed correct) sorted list:

    10 15 15 28 30 30 33 40 45 45 45 45 45 46 50 50 50 52 55 55 55 58 60 60 65 65 65 65 65 65 65 65 68 68 70 70 70 70 70 73 75 75 75 75 75 75 78 80 80 80 82 85 85 85 90 90 90 90 95 95 95 99 100 105 115 120 120 125 125 137 140 145 200

    As shown in the image that I posted, Webassign said that 10 (lower limit) and 38 (10+28) were correct for the first class. But how do I calculate the next lower limit for the second class? I assumed that it was 48, but that now doesn't make sense because then it excludes 39-47.

    Would it be 40-68? Then 70-98? etc?

    Thanks!
     
  12. Jan 28, 2013 #11

    Ray Vickson

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    Why can't you just take the classes to be 10--38, 38(+0)--66, 66(+0)--94, etc? Or, if you happen to get class boundaries exactly on some data points, shift the boundaries to, for example, (9.5,37.5),(37.5,65.5), etc.
     
  13. Jan 28, 2013 #12
    Wait, sorry, it said that 38 wasn't correct for the upper limit. Is there a reason that the answer wouldn't be 38? Because that now makes perfect sense that it would be 10-38,38-66, etc. :/
     
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