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help123please.

  • one year ago

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

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

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

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

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

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

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

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

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

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

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

  12. pooja195
    • one year ago
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    hold on

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

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

  15. pooja195
    • one year ago
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    have u done that

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

  17. help123please.
    • one year ago
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    Median: 68

  18. pooja195
    • one year ago
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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

  19. help123please.
    • one year ago
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    hmmm.

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

  21. pooja195
    • one year ago
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    srry

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

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

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

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

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

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

  28. help123please.
    • one year ago
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    Please.

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

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

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

  32. help123please.
    • one year ago
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    @eigenschmeigen

  33. pooja195
    • one year ago
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    @jim_thompson5910

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

  35. help123please.
    • one year ago
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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

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

  37. jim_thompson5910
    • one year ago
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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

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

  39. help123please.
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    ok.....hmmm...

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

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

  42. help123please.
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    so 4 sets of data in the lower quartile and 4 in the upper?

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

  44. help123please.
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    Okay.

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

  46. jim_thompson5910
    • one year ago
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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

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

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

  49. help123please.
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    Okay.

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

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

  52. jim_thompson5910
    • one year ago
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    ok

  53. help123please.
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    70.5

  54. jim_thompson5910
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    good

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

  56. help123please.
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    i don't know how to get that...

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

  58. jim_thompson5910
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    it's basically the distance from each endpoint of the box

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

  60. help123please.
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    70.5-64.5?

  61. jim_thompson5910
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    yep

  62. help123please.
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    thats 6.

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

  64. help123please.
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    The min and max?

  65. jim_thompson5910
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    yes, the smallest and largest data values

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

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

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

  69. help123please.
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    Why 1.5?

  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

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

  72. help123please.
    • one year ago
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    Oh.

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

  74. jim_thompson5910
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    and people agree on it

  75. help123please.
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    and the iqr is?

  76. jim_thompson5910
    • one year ago
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    you just found it

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

  78. jim_thompson5910
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    yep

  79. help123please.
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    1.5 x 6 = 9

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

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

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

  83. help123please.
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    56.

  84. jim_thompson5910
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    no

  85. jim_thompson5910
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    68 - 9 is not 56

  86. jim_thompson5910
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    it's 59

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

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

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

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

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

  92. jim_thompson5910
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    any smaller than 59, and the min is an outlier

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

  94. jim_thompson5910
    • one year ago
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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

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

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

  97. help123please.
    • one year ago
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    ill brb

  98. jim_thompson5910
    • one year ago
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    ok

  99. help123please.
    • one year ago
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    im back

  100. jim_thompson5910
    • one year ago
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    ok

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

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

  103. jim_thompson5910
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    68

  104. help123please.
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    68. so 68-9... 59?

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

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

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

  108. help123please.
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    Oh...Ok.

  109. help123please.
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    68+1.5*9

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

  111. help123please.
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    right,sorry

  112. help123please.
    • one year ago
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    77

  113. jim_thompson5910
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    good, is the max smaller or larger than 77?

  114. help123please.
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    The max is 73, so its larger.

  115. jim_thompson5910
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    you mean, the max is smaller than 77?

  116. help123please.
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    Right.

  117. jim_thompson5910
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    so does that make the max an outlier?

  118. help123please.
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    no.

  119. help123please.
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    wait..

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

  121. jim_thompson5910
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    you're correct

  122. help123please.
    • one year ago
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    nvm.

  123. help123please.
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    oh,okay.

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

  125. help123please.
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    Alright,thanks!

  126. jim_thompson5910
    • one year ago
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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|

  127. help123please.
    • one year ago
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    Yes.

  128. jim_thompson5910
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    |dw:1363736267701:dw|

  129. jim_thompson5910
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    this is if you don't have any outliers

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

  131. jim_thompson5910
    • one year ago
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    yw

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