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Math Help - Can someone check my box and whisker plot for me

  1. #1
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    Can someone check my box and whisker plot for me

    I haven't made one in a long time and just want to make sure it's correct. The .pdf is the actual graph I made. It's not to scale or anything. I'm more interested in whether the numbers are correct. The .pdf file is the data set and how I calculated the numbers. If someone knows a way of creating a box and whisker plot using Excel, I'd love to learn how. I've tried looking online in the past but haven't really found anything.

    I didn't label it but there are fifty values in the data set.
    Attached Thumbnails Attached Thumbnails Can someone check my box and whisker plot for me-box-whisker.pdf   Can someone check my box and whisker plot for me-box-plot-data.pdf  
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  2. #2
    Junior Member RHandford's Avatar
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    Hi

    I have not checked you answers yet, but in the mean time I hope this may help you.
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  3. #3
    Junior Member RHandford's Avatar
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    Hi

    I seem to get slightly different answers to you.

    Can someone check my box and whisker plot for me-screenshot.png
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  4. #4
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    Sorry that it's taken me so long to respond. I should have posted the manner in which I got all of my numbers. Anyway, here goes:

    Q_{1}=0.25(50)=12.5=13th value in the ordered data set=7
    Q_{2}(median)=0.5(50)=25=25th value in the ordered data set=10
    Q_{3}=0.75(50)=37.5=38th value in the ordered data set=16[/tex] (I messed up on this one)

    Now, for the upper and lower whiskers, I was taught to take the interquartile range (IQR) Q_{3}-Q_{1}, multiply it by 1.5 and then add the product to the Q_{3} to find the upper whisker and subtract the product from [tex]Q_{1} to find the lower whisker. So that would be:

    IQR=Q_{3}-Q_{1}=16-7=9
    Lower Whisker=Q_{1}-1.5(IQR)=-5.65 But since the lowest value in the data set is 3, 3 becomes our lower whisker. I screwed up on that too.
    Upper Whisker=Q_{3}+1.5(IQR)=29.5Upper whisker is 29.5. There is one value (32) higher than the upper whisker. That value is an outlier.
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