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    You use 0.5 as well as 39.5.
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    (Original post by Air1337)
    You use 0.5 as well as 39.5.

    I'm confused on this as well, but for the frequency of '3' do you use the 0.5 value or the 9.5 value?
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    (Original post by Secret.)
    I'm confused on this as well, but for the frequency of '3' do you use the 0.5 value or the 9.5 value?
    Is this to find the median? And do you use the b + (n/2 - f)/fc x class width method for it?

    I don't understand if you're referring to finding the median or not. If the median was in that class then b would be 0.5. If the median was in the frequency of 6 class then you would use 9.5 as the b value.
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    (Original post by Air1337)
    Is this to find the median? And do you use the b + (n/2 - f)/fc x class width method for it?

    I don't understand if you're referring to finding the median or not. If the median was in that class then b would be 0.5. If the median was in the frequency of 6 class then you would use 9.5 as the b value.
    No just asking which numbers go with which for the 3 do we use the 0.5 value?
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    (Original post by Secret.)
    No just asking which numbers go with which for the 3 do we use the 0.5 value?
    For the frequency 3, the class boundaries are 0.5 and 9.5, assuming the data was rounded to the nearest mm.

    So, both values are associated with it.
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    (Original post by Secret.)
    No just asking which numbers go with which for the 3 do we use the 0.5 value?
    As ghostwalker said both values are associated with it. For the class with frequency 7 you would use 19.5 and 29.5 making the class width 10. If the first class had 0 - 9 you would simply use 0 and 9.5.
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    (Original post by ghostwalker)
    For the frequency 3, the class boundaries are 0.5 and 9.5, assuming the data was rounded to the nearest mm.

    So, both values are associated with it.
    (Original post by Air1337)
    As ghostwalker said both values are associated with it. For the class with frequency 7 you would use 19.5 and 29.5 making the class width 10. If the first class had 0 - 9 you would simply use 0 and 9.5.

    But if I had to use interpolation and the frequency was 7 which value would I use 19.5 or 29.5?
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    (Original post by Secret.)
    But if I had to use interpolation and the frequency was 7 which value would I use 19.5 or 29.5?
    You would need both values.

    Say you wanted the 4th value in that class, then with interpolation you'd use:

    19.5 + (29.5-19.5)*4/7
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    (Original post by ghostwalker)
    You would need both values.

    Say you wanted the 4th value in that class, then with interpolation you'd use:

    19.5 + (29.5-19.5)*4/7

    Wouldn't we use the 9.5? As we have 3 and then

    So if we wanted a frequency of 5 would we do

    9.5 + (2/6)*10

    :O?
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    (Original post by Secret.)
    Wouldn't we use the 9.5? As we have 3 and then

    So if we wanted a frequency of 5 would we do

    9.5 + (2/6)*10

    :O?
    Sorry, I don't understand you. The frequency of the class is fixed by the data, it can't change. Can you clarify what you're asking; as much detail as possible please.
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    (Original post by ghostwalker)
    Sorry, I don't understand you. The frequency of the class is fixed by the data, it can't change. Can you clarify what you're asking; as much detail as possible please.

    If we had to interpolate and for example we wanted the median, it will be in the 10-19 class so would we do


    9.5 + (\frac{1}{7}*10) to give us 10.928...?
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    (Original post by Secret.)
    If we had to interpolate and for example we wanted the median, it will be in the 10-19 class so would we do


    9.5 + (\frac{1}{7}*10) to give us 10.928...?
    OK. I understand you now.

    Which entry do you consider to be the median? Since the sum of the frequencies is 20, are you going for the 10th, or the midpoint of 10th to 11th?

    Edit:

    I'll assume it's the 10th.

    The 10th entry is in the first one in the third class with boundaries 19.5 and 29.5

    So, with interpolation, it will be 19.5 + (1/7)(29.5-19.5)

    You wouldn't use the 9.5 at all.
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    (Original post by ghostwalker)
    OK. I understand you now.

    Which entry do you consider to be the median? Since the sum of the frequencies is 20, are you going for the 10th, or the midpoint of 10th to 11th?

    Edit:

    I'll assume it's the 10th.

    The 10th entry is in the first one in the third class with boundaries 19.5 and 29.5

    So, with interpolation, it will be 19.5 + (1/7)(29.5-19.5)

    You wouldn't use the 9.5 at all.

    Aah okay, I get confused as to which number to pick the 9.5 or 19.5 how do you know which one it is? Or should I e.g. in this case the the max height is 39 so the median will be around 20 so it would be the 19.5 value? Or is there a way of knowing which one? Thanks!
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    (Original post by Secret.)
    Aah okay, I get confused as to which number to pick the 9.5 or 19.5 how do you know which one it is? Or should I e.g. in this case the the max height is 39 so the median will be around 20 so it would be the 19.5 value? Or is there a way of knowing which one? Thanks!
    To find the median, you have to find the middle entry if they are all put in ascending order.

    e.g. The median of 1,1,4,5,5 is 4, and the median of 1,1,4,1000,5000 is 4.

    The median is the value of that middle entry.

    With grouped data here, you find that there are (3+6+7+4=) 20 entries, so it's going to be the value of 20/2 = 10th entry (note there are slightly different methods for this, but go with what you're taught).

    There are 3 entries in the first class, 6 in the second, so the 10th entry must be in the third class, and the lower bound of the third class is 19.5 and that is why you choose 19.5.

    You're interpolating between 19.5 and 29.5.
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    (Original post by ghostwalker)
    To find the median, you have to find the middle entry if they are all put in ascending order.

    e.g. The median of 1,1,4,5,5 is 4, and the median of 1,1,4,1000,5000 is 4.

    The median is the value of that middle entry.

    With grouped data here, you find that there are (3+6+7+4=) 20 entries, so it's going to be the value of 20/2 = 10th entry (note there are slightly different methods for this, but go with what you're taught).

    There are 3 entries in the first class, 6 in the second, so the 10th entry must be in the third class, and the lower bound of the third class is 19.5 and that is why you choose 19.5.

    You're interpolating between 19.5 and 29.5.

    Aah right! Thank you very much

    Wont let me rep you again but thanks
 
 
 
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