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Standard deviation and interquartile range

  1. Feb 17, 2012 #1
    1. The problem statement, all variables and given/known data
    There are 10 observations.
    (a) Suppose the IQR is 8. Is it possible that the standard deviation is 1?
    (b) Suppose the standard deviation is 8. Is it possible that the IQR is 1?


    2. Relevant equations
    IQR = R3 - R1
    91ea0949a71eb719b5626e83bd3d6d4a.png

    3. The attempt at a solution
    My argument is that (a) is impossible because the variation of the middle fifty percent of the data (IQR) cannot be greater than the variation of the entire data (standard deviation).
    However, I'm unsure of whether I have to use the fact that there are 10 observations..
     
  2. jcsd
  3. Feb 17, 2012 #2

    Bacle2

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    Note that the standard deviation does not really describe the variation of the entire data; in many cases, the sd is not a good measure of variability of the data -- and this is why you use the IQR. So when you use a 5-number summary it is usually because your data is either not symmetrically-distributed and has outliers.
     
  4. Feb 17, 2012 #3
    Okay. So the IQR tells the variation of the middle fifty percent of the data, but the sd is less reliable if there are outliers. Does this mean that the IQR and sd don't have a relationship, since sd is not resistant? Would both cases be possible, then?
     
  5. Feb 18, 2012 #4

    Bacle2

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    Sorry, I misread your problem; I'll get back to you by tomorrow.
     
  6. Feb 19, 2012 #5

    Bacle2

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    Hi, Sorry for the delay. Sorry, I have not been able to solve this yet. Let me assume we have 8 values {a1,...,a8} ,ordered by size, i.e., i<j →ai ≤ aj, to deal more easily with the quartiles; I think the results here will generalize:

    This is for b, i.e, sd=8, and IQR=1.

    I think thisis possible: minimize variability by having all values below the 75%-ile one unit
    below it, then the rest all equal, e.g:

    1,1,1,1,1,2,2,2,2,27

    which gives you a standard deviation of 8.1. Then you can decrease the largest value so that your deviation is 8.

    The problem I see with your argument is that variability of the data is measured against the mean, not between different values.

    I'm still working on the other one.
    See: http://easycalculation.com/statistics/standard-deviation.php
     
    Last edited: Feb 19, 2012
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