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Mean with standard deviation or Median with IQR?

  1. Nov 21, 2013 #1

    3vo

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    Hi guys,

    I hope someone is able to help me with this, I'm currently stuck on a problem.

    1. I was given some data (in continuous, grouped form) regarding phone call times for a call center agent and asked to represent the data using the most accurate form of average.
    I initially calculated an estimate of the median and IQR using interpolation, followed by estimation of the mean with standard deviation using midpoints for each group.

    The answers however were very different.
    Median 3.44 with IQR 9.5
    Mean 8.6 with standard deviation of 14.5


    I now need to justify which is the better representation of the data, mean or median?
    My ogive graph seems to indicate that the distribution for data is wide and uneven and based on this I was always under the impression that if the distribution is skewed it is better to use median as it is not affected by outliers. However I was informed by someone that the mean was the more accurate representation in this case. Any ideas why this may be?

    Can anyone explain to me what to do with standard deviation value or IQR. I understand they are both measures of spread, but what do they mean to the data? All my text books seem to keep reiterating that they are measures of spread without explaining what to do with them regarding accuracy


    2. Below is a copy of the data table I've complied
    t group mid(x) f c.f fx fx2
    0 ≤ t < 2 1 80 80 80 80
    2 ≤ t < 4 3 53 133 159 477
    4 ≤ t < 6 5 19 152 95 475
    6 ≤ t < 10 8 22 174 176 1408
    10 ≤ t < 20 15 31 205 465 6975
    20 ≤ t < 30 25 16 221 400 10000
    30 ≤ t < 60 45 15 236 675 30375
    Total - 236 236 2050 49790


    3. I believe that due to the uneven distribution that the median may be the better representation for this data, however I have also noticed that the spread is very wide and understand that median is more to do with the central tendency. I'm unsure if this would then disregards the high freq in the first group and may explain why median is not the best representation in the case of wide distributions rather than just uneven?
     

    Attached Files:

  2. jcsd
  3. Nov 21, 2013 #2

    haruspex

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    IMHO, it makes no sense to ask what is the best way to represent data without first understanding how the representation will be used to make decisions.
     
  4. Nov 22, 2013 #3

    3vo

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    Hi Haruspex,

    Thank you for your reply.

    The main part of my assignment brief was to show I was able to calculate an estimate for both mean and median with measures of spread.

    However for the final part I only need to justify which of the two averages is the better representation for this data as a whole. No further conclusions or decisions would be made or required from this data.

    I'm now stuck on which of two measures best represents this data set.

    From my understanding (and please correct me if I am wrong) is that 68% of the values are less then one SD from the mean value. And 95% are less then 2 SD.

    Looking at my cumulative frequency graph I can see that most of the data does fall within one SD from the mean value of 8. However everything I've read either online or in my text book also seems to suggest if the distribution is ever uneven, to always use median. I've also calculated that outliers are present after the 25.25 value and this would normally affect the mean value. Would it also have an impact on SD or is SD resistant to outliers?

    I understand the median gives a better idea of central tendency than mean and is resistant to presence of outliers, would this be enough justification to use median as a better representation than mean?
     
    Last edited: Nov 22, 2013
  5. Nov 22, 2013 #4

    haruspex

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