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help123please. Group Title

Display the data in the chart attached below by making a box and whisker plot. Explain how you made it.

  • one year ago
  • one year ago

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  1. help123please. Group Title
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    • one year ago
  2. help123please. Group Title
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    @rajathsbhat

    • one year ago
  3. rajathsbhat Group Title
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    statistics....we meet again.

    • one year ago
  4. help123please. Group Title
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    LOL. Are you not so good with them?

    • one year ago
  5. rajathsbhat Group Title
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    not so good is an overstatement.

    • one year ago
  6. rajathsbhat Group Title
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    *understatement. I suck at it.

    • one year ago
  7. help123please. Group Title
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    Oh, ok. thats fine, no worries. Ill go beg someone else. :D

    • one year ago
  8. help123please. Group Title
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    Thanks for looking,tho.

    • one year ago
  9. pooja195 Group Title
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    http://www.purplemath.com/modules/boxwhisk3.htm

    • one year ago
  10. pooja195 Group Title
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    see if this helps :)

    • one year ago
  11. help123please. Group Title
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    Hmm... could you explain it step by step?

    • one year ago
  12. pooja195 Group Title
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    hold on

    • one year ago
  13. help123please. Group Title
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    ok. ty:)

    • one year ago
  14. pooja195 Group Title
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    you find the median of your data.

    • one year ago
  15. pooja195 Group Title
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    have u done that

    • one year ago
  16. help123please. Group Title
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    one second - let me do it.

    • one year ago
  17. help123please. Group Title
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    Median: 68

    • one year ago
  18. pooja195 Group Title
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    goodTo divide the data into quarters, you then find the medians of these two halves. Note: If you have an even number of values, so the first median was the average of the two middle values, then you include the middle values in your sub-median computations. If you have an odd number of values, so the first median was an actual data point, then you do not include that value in your sub-median computations. That is, to find the sub-medians, you're only looking at the values that haven't yet been used

    • one year ago
  19. help123please. Group Title
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    hmmm.

    • one year ago
  20. help123please. Group Title
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    let me read all that.

    • one year ago
  21. pooja195 Group Title
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    srry

    • one year ago
  22. help123please. Group Title
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    why are you sorry?

    • one year ago
  23. pooja195 Group Title
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    too much to read :)

    • one year ago
  24. help123please. Group Title
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    indeed. lol.

    • one year ago
  25. help123please. Group Title
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    could u simplify it?

    • one year ago
  26. pooja195 Group Title
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    divide the data into quarters, you then find the medians of these two halves.

    • one year ago
  27. help123please. Group Title
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    hmmm...could u show me the steps by using the DRAW button?

    • one year ago
  28. help123please. Group Title
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    Please.

    • one year ago
  29. pooja195 Group Title
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    im not surehttp://www.purplemath.com/modules/boxwhisk.htm

    • one year ago
  30. help123please. Group Title
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    Hmm. Okay. Thanks.

    • one year ago
  31. help123please. Group Title
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    Lets see.....

    • one year ago
  32. help123please. Group Title
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    @eigenschmeigen

    • one year ago
  33. pooja195 Group Title
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    @jim_thompson5910

    • one year ago
  34. jim_thompson5910 Group Title
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    what's your question?

    • one year ago
  35. help123please. Group Title
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    I have to make a box and whisker plot for the data shown at http://assets.openstudy.com/updates/attachments/5148e6dee4b05e69bfab6578-help123please.-1363732200597-screenshot20130319at5.28.53pm.png

    • one year ago
  36. help123please. Group Title
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    I also would like to know how to make box and whisker plots by myself.

    • one year ago
  37. jim_thompson5910 Group Title
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    Find the median first Then split the data into two groups (the median is the cut off point) and find the median of each group. The first quartile is the median of the lower half, the third quartile is the median of the upper half

    • one year ago
  38. help123please. Group Title
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    the median is 68.

    • one year ago
  39. help123please. Group Title
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    ok.....hmmm...

    • one year ago
  40. help123please. Group Title
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    oohhhh divide it into 2?

    • one year ago
  41. help123please. Group Title
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    8/2=4

    • one year ago
  42. help123please. Group Title
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    so 4 sets of data in the lower quartile and 4 in the upper?

    • one year ago
  43. jim_thompson5910 Group Title
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    yes the median is 68

    • one year ago
  44. help123please. Group Title
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    Okay.

    • one year ago
  45. help123please. Group Title
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    show me step by step how to make it using the DRAW button please?

    • one year ago
  46. jim_thompson5910 Group Title
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    so the sorted number set 61, 64, 65, 68, 68, 70, 71, 73 turns into 61, 64, 65, 68 | 68, 70, 71, 73 when you split the data in half along the median

    • one year ago
  47. help123please. Group Title
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    Uh-huh, I see.

    • one year ago
  48. jim_thompson5910 Group Title
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    the median of 61, 64, 65, 68 is 64.5, so Q1 = 64.5

    • one year ago
  49. help123please. Group Title
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    Okay.

    • one year ago
  50. jim_thompson5910 Group Title
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    what's Q3?

    • one year ago
  51. help123please. Group Title
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    Let me see. One second.

    • one year ago
  52. jim_thompson5910 Group Title
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    ok

    • one year ago
  53. help123please. Group Title
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    70.5

    • one year ago
  54. jim_thompson5910 Group Title
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    good

    • one year ago
  55. jim_thompson5910 Group Title
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    what's your IQR

    • one year ago
  56. help123please. Group Title
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    i don't know how to get that...

    • one year ago
  57. jim_thompson5910 Group Title
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    IQR = Q3 - Q1

    • one year ago
  58. jim_thompson5910 Group Title
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    it's basically the distance from each endpoint of the box

    • one year ago
  59. help123please. Group Title
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    Oh,ok.

    • one year ago
  60. help123please. Group Title
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    70.5-64.5?

    • one year ago
  61. jim_thompson5910 Group Title
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    yep

    • one year ago
  62. help123please. Group Title
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    thats 6.

    • one year ago
  63. jim_thompson5910 Group Title
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    now find the min and max

    • one year ago
  64. help123please. Group Title
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    The min and max?

    • one year ago
  65. jim_thompson5910 Group Title
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    yes, the smallest and largest data values

    • one year ago
  66. help123please. Group Title
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    oh... U mean min = 61 max = 73

    • one year ago
  67. jim_thompson5910 Group Title
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    good, now determine if the min is an outlier or not

    • one year ago
  68. jim_thompson5910 Group Title
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    you do this by seeing if the min is above median - 1.5*IQR

    • one year ago
  69. help123please. Group Title
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    Why 1.5?

    • one year ago
  70. jim_thompson5910 Group Title
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    just a number that some statistician a long time ago determined was the best number to see if outliers exist on a box and whisker plot

    • one year ago
  71. jim_thompson5910 Group Title
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    I think it was Tukey

    • one year ago
  72. help123please. Group Title
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    Oh.

    • one year ago
  73. jim_thompson5910 Group Title
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    I'm not sure how or why he got 1.5, but it works

    • one year ago
  74. jim_thompson5910 Group Title
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    and people agree on it

    • one year ago
  75. help123please. Group Title
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    and the iqr is?

    • one year ago
  76. jim_thompson5910 Group Title
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    you just found it

    • one year ago
  77. help123please. Group Title
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    o sorry....6

    • one year ago
  78. jim_thompson5910 Group Title
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    yep

    • one year ago
  79. help123please. Group Title
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    1.5 x 6 = 9

    • one year ago
  80. jim_thompson5910 Group Title
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    subtract that from the median 68

    • one year ago
  81. help123please. Group Title
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    Hmm. Ok. WOW, lots of steps :/ LOL.

    • one year ago
  82. jim_thompson5910 Group Title
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    it's not so bad once you get used to them

    • one year ago
  83. help123please. Group Title
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    56.

    • one year ago
  84. jim_thompson5910 Group Title
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    no

    • one year ago
  85. jim_thompson5910 Group Title
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    68 - 9 is not 56

    • one year ago
  86. jim_thompson5910 Group Title
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    it's 59

    • one year ago
  87. help123please. Group Title
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    o right sorry.

    • one year ago
  88. jim_thompson5910 Group Title
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    so that's your lowest threshold

    • one year ago
  89. help123please. Group Title
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    i punched the wrong numbers into the calculator.

    • one year ago
  90. jim_thompson5910 Group Title
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    basically that's the smallest the min can be

    • one year ago
  91. help123please. Group Title
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    sorry bout that.

    • one year ago
  92. jim_thompson5910 Group Title
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    any smaller than 59, and the min is an outlier

    • one year ago
  93. help123please. Group Title
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    Ah,ok.

    • one year ago
  94. jim_thompson5910 Group Title
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    so we do NOT have an outlier for the min because the min (which is 61) is greater than the threshold of 59

    • one year ago
  95. jim_thompson5910 Group Title
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    now compute median + 1.5*IQR

    • one year ago
  96. help123please. Group Title
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    Mhmm. Okay. One second - will log off for like 2 mins,k?

    • one year ago
  97. help123please. Group Title
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    ill brb

    • one year ago
  98. jim_thompson5910 Group Title
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    ok

    • one year ago
  99. help123please. Group Title
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    im back

    • one year ago
  100. jim_thompson5910 Group Title
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    ok

    • one year ago
  101. jim_thompson5910 Group Title
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    did you find median + 1.5*IQR

    • one year ago
  102. help123please. Group Title
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    lets see. the median was?

    • one year ago
  103. jim_thompson5910 Group Title
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    68

    • one year ago
  104. help123please. Group Title
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    68. so 68-9... 59?

    • one year ago
  105. jim_thompson5910 Group Title
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    you already did that one

    • one year ago
  106. help123please. Group Title
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    its not the answer?

    • one year ago
  107. jim_thompson5910 Group Title
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    no I wanted you to find median + 1.5*IQR

    • one year ago
  108. help123please. Group Title
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    Oh...Ok.

    • one year ago
  109. help123please. Group Title
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    68+1.5*9

    • one year ago
  110. jim_thompson5910 Group Title
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    yeah but the IQR is 6, not 9

    • one year ago
  111. help123please. Group Title
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    right,sorry

    • one year ago
  112. help123please. Group Title
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    77

    • one year ago
  113. jim_thompson5910 Group Title
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    good, is the max smaller or larger than 77?

    • one year ago
  114. help123please. Group Title
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    The max is 73, so its larger.

    • one year ago
  115. jim_thompson5910 Group Title
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    you mean, the max is smaller than 77?

    • one year ago
  116. help123please. Group Title
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    Right.

    • one year ago
  117. jim_thompson5910 Group Title
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    so does that make the max an outlier?

    • one year ago
  118. help123please. Group Title
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    no.

    • one year ago
  119. help123please. Group Title
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    wait..

    • one year ago
  120. jim_thompson5910 Group Title
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    so there are no outliers

    • one year ago
  121. jim_thompson5910 Group Title
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    you're correct

    • one year ago
  122. help123please. Group Title
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    nvm.

    • one year ago
  123. help123please. Group Title
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    oh,okay.

    • one year ago
  124. jim_thompson5910 Group Title
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    so you have enough info to plot the box and whisker plot

    • one year ago
  125. help123please. Group Title
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    Alright,thanks!

    • one year ago
  126. jim_thompson5910 Group Title
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    The basic format (the chances of your box and whisker plot looking exactly like this are very slim) is this |dw:1363736245038:dw|

    • one year ago
  127. help123please. Group Title
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    Yes.

    • one year ago
  128. jim_thompson5910 Group Title
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    |dw:1363736267701:dw|

    • one year ago
  129. jim_thompson5910 Group Title
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    this is if you don't have any outliers

    • one year ago
  130. help123please. Group Title
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    Okay. I think I understand a bit better- thanks!

    • one year ago
  131. jim_thompson5910 Group Title
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    yw

    • one year ago
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