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anonymous

  • one year ago

The data set gives the number of bottles filled by each of the workers in a bottling plant in one day. {36, 18, 16, 28, 68, 35, 37, 66, 38, 40, 41, 44, 72, 29} The best measure of center for this data set is the , and its value expressed up to one decimal place is

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  1. anonymous
    • one year ago
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    The best would be the median

  2. anonymous
    • one year ago
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    how do I find the value?

  3. welshfella
    • one year ago
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    arrange the values in ascending order the median is the middle number (if the total number is odd) or he mean of the 2 middle numbers.

  4. anonymous
    • one year ago
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    @welshfella I know that part

  5. anonymous
    • one year ago
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    I need to know the value

  6. welshfella
    • one year ago
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    i've just explained how to get it.

  7. welshfella
    • one year ago
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    there are 2 parts to the question 1. what is the best measure of center 2. what is the value of this best measure.

  8. welshfella
    • one year ago
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    OK?

  9. anonymous
    • one year ago
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    order these numbers; {36, 18, 16, 28, 68, 35, 37, 66, 38, 40, 41, 44, 72, 29} , then find the middle number of the order

  10. IrishBoy123
    • one year ago
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    out of interest, why the median and not the mean? https://gyazo.com/adeb20b7894f04112ea381c640e00bba is this really that skewed?

  11. anonymous
    • one year ago
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    mean is affected by outliers

  12. IrishBoy123
    • one year ago
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    yes it is, but the trendline suggests that there aren't really any outliers. i reason that the go-to measure of central tendency is the arithmetic mean. now, if you work total output as 14 times mean for this sample vs 14 times median, you get very different results. in fact you are out by nearly 10% if you use median. this is not a massively skewed distribution. skewness \(\frac{\mu - \nu}{\sigma} = 0.18\). not totally sure how to interpret that, but it certainly isn't a big number.

  13. IrishBoy123
    • one year ago
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    i'm just wondering why the median was selected as best without any analysis, and the whole thread's focus is on how you calc a median. if you take the outlier argument too seriously, you never get to use the arithmetic mean unless every data point is exactly the same :p

  14. welshfella
    • one year ago
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    you have a valid point there @irishboy. I guess this is a 'borderline' case.

  15. IrishBoy123
    • one year ago
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    thanks @welshfella ! i can see it both ways, i just wonder is it really about judgement or is there a test? i do see those 3 higher numbers but they are 3 in a set of 14. if it was just one, i might see it differently.

  16. welshfella
    • one year ago
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    yes I agree about the outliers I wonder if people lean towards the median if the amount of data is small because its easier to calculate!?

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