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Statistics 1. please help me understand this!! watch

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    Edexcel S1. chapter 4. EX 4G question 7.

    Question:The labelling on bags of garden compost indicates that the bags weigh 20 kg.
    The weights of a random sample of 50 bags are summarised in the table opposite.Using linear interpolation, estimate the median for below:


    Weight in kg Frequency
    14.6–14.8 ] 1
    14.8–18.0 ] 0
    18.0–18.5 ] 5
    18.5–20.0 ] 6
    20.0–20.2 ] 22
    20.2–20.4 ] 15
    20.4–21.0 ] 1


    Edexcel S1. chapter 4. EX 4G question 7.
    ANSWER:

    median=20+( 13/22) ×0.2=20.118181818=20.118

    Can you please explain to me what is LINEAR INTERPOLATION? I can not find it in my textbook, or perhaps I may have done it without knowing it is called 'linear interpolation'. (btw, i am generally quick to understand!)
    I do not understand how they got the answer or the calculation they did???
    where did they get the number 20 from? AND 13/22???

    Can anyone please please help me out.. I am really stuck.
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    Over 21 veiws and yet no reply!!
    Come on people...please be helpful!?

    Help me::?
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    They seem to have used median position = n/2 = 50/2= 25th value...which lies in the group 20.0-20.2 (called median class).
    Some S1 mark schemes also use (n+1)/2.

    Since the freqs 1,0,5,6 sum to 12...this means the 25th value is actually the 13th value in the group 20.0-20.2.
    0.2 = class width and 13/22 is the fraction into this group you have to go into. The 20 is called the lower class boundary (of the median class).

    Does this help?
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    Linear interpolation is estimating an intermediate value from 2 given values.

    So say that when x=5 y=12 and when x=9 y=15, then we might guess that when x=7, y=(15+12)/2 = 13.5.

    Does that make sense?
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    (Original post by pHneutral)
    Edexcel S1. chapter 4. EX 4G question 7.

    Question:The labelling on bags of garden compost indicates that the bags weigh 20 kg.
    The weights of a random sample of 50 bags are summarised in the table opposite.Using linear interpolation, estimate the median for below:


    Weight in kg Frequency
    14.6–14.8 ] 1
    14.8–18.0 ] 0
    18.0–18.5 ] 5
    18.5–20.0 ] 6
    20.0–20.2 ] 22
    20.2–20.4 ] 15
    20.4–21.0 ] 1

    Edexcel S1. chapter 4. EX 4G question 7.
    ANSWER:

    median=20+( 13/22) ×0.2=20.118181818=20.118

    Can you please explain to me what is LINEAR INTERPOLATION? I can not find it in my textbook, or perhaps I may have done it without knowing it is called 'linear interpolation'. (btw, i am generally quick to understand!)
    I do not understand how they got the answer or the calculation they did???
    where did they get the number 20 from? AND 13/22???

    Can anyone please please help me out.. I am really stuck.
    Ok so I reckon you are supposed to draw a cumulative frequency graph and join up the points and then work out te median using the line. If you were to EXtrapolate, you would extend the line beyond your range of values (this is bad and unrliable) so INterpolating means estimating WITHIN your values. Make sense?
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    (Original post by pHneutral)
    Edexcel S1. chapter 4. EX 4G question 7.

    Question:The labelling on bags of garden compost indicates that the bags weigh 20 kg.
    The weights of a random sample of 50 bags are summarised in the table opposite.Using linear interpolation, estimate the median for below:


    Weight in kg Frequency
    14.6–14.8 ] 1
    14.8–18.0 ] 0
    18.0–18.5 ] 5
    18.5–20.0 ] 6
    20.0–20.2 ] 22
    20.2–20.4 ] 15
    20.4–21.0 ] 1


    Edexcel S1. chapter 4. EX 4G question 7.
    ANSWER:

    median=20+( 13/22) ×0.2=20.118181818=20.118

    Can you please explain to me what is LINEAR INTERPOLATION? I can not find it in my textbook, or perhaps I may have done it without knowing it is called 'linear interpolation'. (btw, i am generally quick to understand!)
    I do not understand how they got the answer or the calculation they did???
    where did they get the number 20 from? AND 13/22???

    Can anyone please please help me out.. I am really stuck.
    Linear Interpolation is covered in chapter 2 of that texbook .
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    (Original post by vc94)
    They seem to have used median position = n/2 = 50/2= 25th value...which lies in the group 20.0-20.2 (called median class).
    Some S1 mark schemes also use (n+1)/2.

    Since the freqs 1,0,5,6 sum to 12...this means the 25th value is actually the 13th value in the group 20.0-20.2.
    0.2 = class width and 13/22 is the fraction into this group you have to go into. The 20 is called the lower class boundary (of the median class).

    Does this help?
    thanks the above is actually very helpful.
    so if i was to try and put the numbers in to words, is below correct??

    Median interpolated= A + (B/C) * D
    A = Lower class boundry of the median class
    B= which is 13 in our case...i am unable to put that into words..?..why is it not 12 when the above frequencies add up to 12?

    C= the frequency of the group in which the median is to be found
    D= class width

    thnaks
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    B= median position subtract the cumulative frequencies above the median class i.e. 25-12=13.

    The rest is cool!
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    (Original post by ViralRiver)
    Linear Interpolation is covered in chapter 2 of that texbook .
    I am not able to find it, could you please tell me where exactly.

    .if you are reffering to page 20-21, that i believe is just interpolation. I understand that, of how you draw the diagram and solve the equation to get the median or whatever you are looking for. I know that method.

    the above question states the method of using particularly linear interpolation. I think i did this in GCSE STATS but cant not remember it now..

    thanks
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    (Original post by vc94)
    B= median position subtract the cumulative frequencies above the median class i.e. 25-12=13.

    The rest is cool!
    Ok, now i get it!
    Thanks a lot

    I will remember that now hoepfully!
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    thanks :

    vc94
    baffled_mathman
    lolo-x
    ViralRiver

    I understand it now,problem solved,
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    haha im just stuck on the exact same problem! ****in hate stats right now
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    Also, on this question there is a part d which says

    3(mean-median)/standard deviation
    Evaluate this coefficient of these data

    so I subbed in the numbers:

    3x(19.8-20.1)/0.963 (which are correct because the answers in the back of the book says that these are correct)

    And I am not getting the correct answer!! Apparently it is-1.06?!

    HELP PLEASE
 
 
 
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