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    So I'm trying to do this question and it should be relatively easy, I just keep getting different answers to the one in the text book. To make it more confusing SolutionBank gets a different answer to what's at the back of the text book as well.

    A hotel is worried about the reliability of its lift. It keeps a weekly record of the number of times it breaks down over 26 weeks.

    Number of breakdowns - Frequency
    0-1 - 18
    2-3 - 7
    4-5 - 1

    b. Use interpolation to estimate the median number of breakdowns.

    The books got 1.08 and SolutionBank gets 0.722. I get 0.94 so I'm literally so confused right now. Any help would be useful.
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    What is your method?
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    make sure you grouped the data correctly. check your arithmetic in your calculations and check your cumulative frequency calculations. also check if you've done the linear interpolation method properly, it's easy to mix up two numbers (medianClass -minclass/maxclass-minclass = medianPos - minCumulative/maxCumulative - minCumulative)

    it would help if you posted a screenshot of your workings.
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    Here is a video I have done on linear interpolation - it goes through each step. Hope this helps!
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    I know this is an old forum but this is for anyone who still needs it. The answer on solution bank isn't complete - they forgot to multiply 13/18 by 1.5.

    0.722 x 1.5 = 1.08

    So just follow the one on solution bank and finish the last step. The key here is that you have got to assume that the width is 0 - 1.5. To be honest I'm still stuck on why this is, since the data isn't "to the nearest week", it is simply after each week.
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    (Original post by alextheowl)
    I know this is an old forum but this is for anyone who still needs it. The answer on solution bank isn't complete - they forgot to multiply 13/18 by 1.5.

    0.722 x 1.5 = 1.08

    So just follow the one on solution bank and finish the last step. The key here is that you have got to assume that the width is 0 - 1.5. To be honest I'm still stuck on why this is, since the data isn't "to the nearest week", it is simply after each week.
    But it's a discrete distribution, so would that still apply?

    (I'm not sure, by the way - I'm just curious.)
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    (Original post by alextheowl)
    I know this is an old forum but this is for anyone who still needs it. The answer on solution bank isn't complete - they forgot to multiply 13/18 by 1.5.

    0.722 x 1.5 = 1.08

    So just follow the one on solution bank and finish the last step. The key here is that you have got to assume that the width is 0 - 1.5. To be honest I'm still stuck on why this is, since the data isn't "to the nearest week", it is simply after each week.
    I disagree with this. The data is discrete so clearly an estimate of the median has to lie between 0 and 1.
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    (Original post by Notnek)
    I disagree with this. The data is discrete so clearly the median has to lie between 0 and 1.
    That is what I thought at first. The book (and kind of solution bank) say 1.08, so I assumed they were correct. Maybe they're both wrong.
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    (Original post by alextheowl)
    That is what I thought at first. The book (and kind of solution bank) say 1.08, so I assumed they were correct. Maybe they're both wrong.
    The solution bank I’m looking at says it’s 0.722.

    Actually since it’s discrete data I would have thought that it should be the 13.5th value and not the 13th.
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    (Original post by Notnek)
    The solution bank I’m looking at says it’s 0.722.

    Actually since it’s discrete data shouldn’t it be the 13.5th value and not the 13th?
    The solution bank forgot to multiply 13/18 by 1.5 to get 1.08. And I'm as confused as you are now, I will ask my teacher on Monday!
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    (Original post by alextheowl)
    The solution bank forgot to multiply 13/18 by 1.5 to get 1.08. And I'm as confused as you are now, I will ask my teacher on Monday!
    No that would be an incorrect method. It’s clear from the data that the median can’t be more than 1.
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    ghostwalker DFranklin

    I’m still not experienced with this kind of thing so do you mind sharing the answer that you get (question in OP) when you have time?

    I don’t know if methods for finding medians vary between courses.
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    (Original post by Notnek)
    ghostwalker DFranklin

    I’m still not experienced with this kind of thing so do you mind sharing the answer that you get (question in OP) when you have time?

    I don’t know if methods for finding medians vary between courses.
    I'm certainly not experienced with this (to the point that I'm not sure I've ever done this). I agree that it seems nonsensical to estimate the median to be greater than 1 given the data provided.

    Tagging Gregorious as the definite stats master of f38...
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    (Original post by Notnek)
    The solution bank I’m looking at says it’s 0.722.

    Actually since it’s discrete data I would have thought that it should be the 13.5th value and not the 13th.
    Also, my book says for grouped continuous data you do n/2, not (n+1)/2, so maybe because it's grouped it's 13?
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    (Original post by Notnek)
    ghostwalker DFranklin

    I’m still not experienced with this kind of thing so do you mind sharing the answer that you get (question in OP) when you have time?

    I don’t know if methods for finding medians vary between courses.
    Since we have grouped data, the median is taken to be the (1/2) nth reading, i.e. the 13th.

    Interpolation gives us 13/18 = 0.7222..., of the way through the interval.

    The $64,000 question is what is the class width, for this discrete data? I would say it's "1", and so 0.722... is the desired value, since the interpolated value must lie between 0 and 1.

    Edit: On further reflection, this would imply that if the median shifted by one place it could conceivably jump from 1 to 2, if it was at the end of the interval, which doesn't seem right. So, perhaps the class interval should be 2. Making the median 1.444...

    Gregorius would be a good person to get on board with this, I think.
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    (Original post by ghostwalker)
    Since we have grouped data, the median is taken to be the (1/2) nth reading, i.e. the 13th.

    Interpolation gives us 13/18 = 0.7222..., of the way through the interval.

    The $64,000 question is what is the class width, for this discrete data? I would say it's "1", and so 0.722... is the desired value, since the interpolated value must lie between 0 and 1.
    Thanks. I’m not used to grouped discrete data so was unsure and there may not be a standard A Level method for this.

    I doubt this would appear in an A Level exam.
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    (Original post by Notnek)
    Thanks. I’m not used to grouped discrete data so was unsure and there may not be a standard A Level method for this.

    I doubt this would appear in an A Level exam.
    I changed my mind and updated my previous post - sorry!
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    (Original post by ghostwalker)
    I changed my mind and updated my previous post - sorry!
    Does it make sense to have a median greater than 1?
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    (Original post by Notnek)
    Does it make sense to have a median greater than 1?
    Nope.

    I'm having third thoughts now. Oh, dear. :getmecoat:
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    (Original post by ghostwalker)
    Nope.

    I'm having third thoughts now. Oh, dear. :getmecoat:
    I’m glad I’m not the only one who got confused by it

    I actually thought you would choose (n+1)/2 because it’s discrete data. But I’m not used to grouped discrete data as I’ve said.
 
 
 
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