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help123please.

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.
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  2. help123please.
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    @rajathsbhat

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

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

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

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

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

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

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

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

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

    • one year ago
  12. pooja195
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    hold on

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

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

    • one year ago
  15. pooja195
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    have u done that

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

    • one year ago
  17. help123please.
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    Median: 68

    • one year ago
  18. pooja195
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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.
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    hmmm.

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

    • one year ago
  21. pooja195
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    srry

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

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

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

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

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

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

    • one year ago
  28. help123please.
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    Please.

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

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

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

    • one year ago
  32. help123please.
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    @eigenschmeigen

    • one year ago
  33. pooja195
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    @jim_thompson5910

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

    • one year ago
  35. help123please.
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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.
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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
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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.
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    the median is 68.

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

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

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

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

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

    • one year ago
  44. help123please.
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    Okay.

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

    • one year ago
  46. jim_thompson5910
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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.
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    Uh-huh, I see.

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

    • one year ago
  49. help123please.
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    Okay.

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

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

    • one year ago
  52. jim_thompson5910
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    ok

    • one year ago
  53. help123please.
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    70.5

    • one year ago
  54. jim_thompson5910
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    good

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

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

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

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

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

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

    • one year ago
  61. jim_thompson5910
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    yep

    • one year ago
  62. help123please.
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    thats 6.

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

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

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

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

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

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

    • one year ago
  69. help123please.
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    Why 1.5?

    • one year ago
  70. jim_thompson5910
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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
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    I think it was Tukey

    • one year ago
  72. help123please.
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    Oh.

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

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

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

    • one year ago
  76. jim_thompson5910
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    you just found it

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

    • one year ago
  78. jim_thompson5910
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    yep

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

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

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

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

    • one year ago
  83. help123please.
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    56.

    • one year ago
  84. jim_thompson5910
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    no

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

    • one year ago
  86. jim_thompson5910
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    it's 59

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

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

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

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

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

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

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

    • one year ago
  94. jim_thompson5910
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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
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    now compute median + 1.5*IQR

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

    • one year ago
  97. help123please.
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    ill brb

    • one year ago
  98. jim_thompson5910
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    ok

    • one year ago
  99. help123please.
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    im back

    • one year ago
  100. jim_thompson5910
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    ok

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

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

    • one year ago
  103. jim_thompson5910
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    68

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

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

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

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

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

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

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

    • one year ago
  111. help123please.
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    right,sorry

    • one year ago
  112. help123please.
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    77

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

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

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

    • one year ago
  116. help123please.
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    Right.

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

    • one year ago
  118. help123please.
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    no.

    • one year ago
  119. help123please.
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    wait..

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

    • one year ago
  121. jim_thompson5910
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    you're correct

    • one year ago
  122. help123please.
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    nvm.

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

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

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

    • one year ago
  126. jim_thompson5910
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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.
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    Yes.

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

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

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

    • one year ago
  131. jim_thompson5910
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    yw

    • one year ago
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