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Thread: Class Interval Frequency Distribution using 10 intervals

  1. #1
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    Class Interval Frequency Distribution using 10 intervals

    Can somebody please help me because I am really lost with this question and how to go about doing it?

    Organize the following data into a class interval frequency distribution using 10 intervals with frequency (f) and relative frequency (rf) columns.

    Data Set: 100, 97, 99, 70, 72, 75, 82, 68, 85, 88, 71, 77, 93, 94, 54, 59, 83, 87, 98, 84, 72, 96, 98, 89, 74, 98, 77, 82, 83, 98, 90, 95, 85, 76, 62, 72, 36, 21, 42, 86, 75, 42, 91, 90, 81, 78, 79, 74, 82, and 98.

    What are the mean, median, and mode of the following data set? Is the distribution positively or negatively skewed? Which of these measures of central tendency would be most appropriate, and why?

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    Where are you stuck? Do you know the definition of mean, median, and mode? If so, you can get started with that.

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    I am stuck on the whole problem. I don't know how to do the problem doing a class interval frequency distribution using 10 intervals, don't know what the mean, median, and mode is or how to come up with those. In response to your question, I know the definition of the mean, median, and mode but don't understand how to come up with the answers. I really don't know where to start and am confused and lost on this problem.

    Quote Originally Posted by jsterkel View Post
    Where are you stuck? Do you know the definition of mean, median, and mode? If so, you can get started with that.

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    Mean is the "average", although many mathmeticians dislike that term.


    So the mean of 14, 19, 8, 11, 8 is

    (14+19+8+11+8)/5

    60/5

    12 is the mean or "average".
    -------------------------------
    Median is the "middle" value,

    In this case, if we arrange them in order 8,8,11,14,19 we see that 11 is in the middle. 11 is the median.

    ------------------------------------
    The mode is the most frequently occurring value in a dataset.

    Therefore, in this case, the mode is 8.



    Now, can you determine this for your data set?

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    Class Interval Frequency Distribution using 10 intervals

    Okay, I came up with the mean and what I did was as follows:
    I arranged the exam scores from lowest to highest. I added all the exam scores and came up with 3, 958.

    Added all the exam scores = 3,958
    Divided sum by number of values = 3, 958 / 50 = 79.16
    Is 79.16 the Mean?

    I got the Mode which is 98 because in this data set it occurs most frequently.

    When there are so many numbers, values, or scores in a data set such as this one has 50. I don't know how to come up with the Median. How do I know what the "middle" value is in a data set that has 50 numbers?

    Quote Originally Posted by jsterkel View Post
    Mean is the "average", although many mathmeticians dislike that term.


    So the mean of 14, 19, 8, 11, 8 is

    (14+19+8+11+8)/5

    60/5

    12 is the mean or "average".
    -------------------------------
    Median is the "middle" value,

    In this case, if we arrange them in order 8,8,11,14,19 we see that 11 is in the middle. 11 is the median.

    ------------------------------------
    The mode is the most frequently occurring value in a dataset.

    Therefore, in this case, the mode is 8.



    Now, can you determine this for your data set?

  6. #6
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    Yes, you are correct.

    The mean is 79.16

    The mode is 98

    In cases where the number of elements in the data set is even, we average the two middle values to determine the median. What are the two values in the middle?




    Note: edited to clarify that the last sentence refers to median.
    Last edited by jsterkel; 10-18-2011 at 11:57 PM.

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    Class Interval Frequency Distributing using 10 intervals

    I don't know what the two middle values are in the middle. How am I supposed to figure this out when there are 50 scores in a set? There could be so much values in the middle so it's kind of hard to say what exactly the two middle values are in the middle. Can you possibly explain this another way? I ask this because to me anything can mean with a data set having 50 scores that mostly any number could be considered the middle value.

    Quote Originally Posted by jsterkel View Post
    Yes, you are correct.

    The mean is 79.16

    The mode is 98

    In cases where the number of elements in the data set is even, we average the two middle values to determine the median. What are the two values in the middle?




    Note: edited to clarify that the last sentence refers to median.

  8. #8
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    Ok, earlier you said "I arranged the exam scores from lowest to highest."

    That's perfect.

    If the number of values is odd, the median will be the single value in the middle. For instance, in the list 1,2,3,4,5 the median is 3, because there are 2 values higher than 3, and 2 values lower than 3.

    If the number of values is even (1,2,3,4,5,6) we have to average the two "in the middle". So we will look at 3 and 4. There are two numbers smaller than 3, and two numbers higher than 4. The average of 3 and 4 is 3.5. So 3.5 is our median.

    Now, your list is longer, but it's basically the same. Since 50 is an even number, we are looking for 2 values, but which two? If we want the two in the middle, that means that there are 48 that we don't want. Half of 48 is 24, so we want 24 numbers smaller than the "two in the middle", and 24 bigger than the "two in the middle".

    The ones you are looking for are the 25th and 26th values in your list.

    Hope this helps.

  9. #9
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    After you have the median, the next step is "Organize the following data into a class interval frequency distribution using 10 intervals with frequency (f) "


    I think this link will do a better job of explaining this than I can.

    http://www.mathsisfun.com/data/frequ...n-grouped.html

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    Class Interval Frequency Distribution using 10 intervals

    The 25th value in the set of numbers I have is 82. The 26th value in the set of numbers that I have is 82. Please let me know if the following is correct for finding the median:
    25th Value is 82, 26th Value is 82.
    Add Both these Values Together: 82 + 82 = 164.
    Divide Answer From Values By 2: 164 / 2 = 82.
    Is 82 the Median for this big long set of numbers? Thanks for all your help and patience with me in regards to these problems.

    Quote Originally Posted by jsterkel View Post
    Ok, earlier you said "I arranged the exam scores from lowest to highest."

    That's perfect.

    If the number of values is odd, the median will be the single value in the middle. For instance, in the list 1,2,3,4,5 the median is 3, because there are 2 values higher than 3, and 2 values lower than 3.

    If the number of values is even (1,2,3,4,5,6) we have to average the two "in the middle". So we will look at 3 and 4. There are two numbers smaller than 3, and two numbers higher than 4. The average of 3 and 4 is 3.5. So 3.5 is our median.

    Now, your list is longer, but it's basically the same. Since 50 is an even number, we are looking for 2 values, but which two? If we want the two in the middle, that means that there are 48 that we don't want. Half of 48 is 24, so we want 24 numbers smaller than the "two in the middle", and 24 bigger than the "two in the middle".

    The ones you are looking for are the 25th and 26th values in your list.

    Hope this helps.

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