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help123please.

  • 2 years 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.
    • 2 years ago
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  2. help123please.
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    @rajathsbhat

  3. rajathsbhat
    • 2 years ago
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    statistics....we meet again.

  4. help123please.
    • 2 years ago
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    LOL. Are you not so good with them?

  5. rajathsbhat
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    not so good is an overstatement.

  6. rajathsbhat
    • 2 years ago
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    *understatement. I suck at it.

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

  8. help123please.
    • 2 years ago
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    Thanks for looking,tho.

  9. pooja195
    • 2 years ago
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    http://www.purplemath.com/modules/boxwhisk3.htm

  10. pooja195
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    see if this helps :)

  11. help123please.
    • 2 years ago
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    Hmm... could you explain it step by step?

  12. pooja195
    • 2 years ago
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    hold on

  13. help123please.
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    ok. ty:)

  14. pooja195
    • 2 years ago
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    you find the median of your data.

  15. pooja195
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    have u done that

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

  17. help123please.
    • 2 years ago
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    Median: 68

  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

  19. help123please.
    • 2 years ago
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    hmmm.

  20. help123please.
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    let me read all that.

  21. pooja195
    • 2 years ago
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    srry

  22. help123please.
    • 2 years ago
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    why are you sorry?

  23. pooja195
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    too much to read :)

  24. help123please.
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    indeed. lol.

  25. help123please.
    • 2 years ago
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    could u simplify it?

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

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

  28. help123please.
    • 2 years ago
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    Please.

  29. pooja195
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    im not surehttp://www.purplemath.com/modules/boxwhisk.htm

  30. help123please.
    • 2 years ago
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    Hmm. Okay. Thanks.

  31. help123please.
    • 2 years ago
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    Lets see.....

  32. help123please.
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    @eigenschmeigen

  33. pooja195
    • 2 years ago
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    @jim_thompson5910

  34. jim_thompson5910
    • 2 years ago
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    what's your question?

  35. help123please.
    • 2 years 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.
    • 2 years ago
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    I also would like to know how to make box and whisker plots by myself.

  37. jim_thompson5910
    • 2 years 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.
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    the median is 68.

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

  40. help123please.
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    oohhhh divide it into 2?

  41. help123please.
    • 2 years 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
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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
    • 2 years 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.
    • 2 years 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
    • 2 years ago
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    what's Q3?

  51. help123please.
    • 2 years ago
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    Let me see. One second.

  52. jim_thompson5910
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    ok

  53. help123please.
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    70.5

  54. jim_thompson5910
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    good

  55. jim_thompson5910
    • 2 years 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
    • 2 years 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.
    • 2 years 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
    • 2 years ago
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    now find the min and max

  64. help123please.
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    The min and max?

  65. jim_thompson5910
    • 2 years ago
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    yes, the smallest and largest data values

  66. help123please.
    • 2 years ago
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    oh... U mean min = 61 max = 73

  67. jim_thompson5910
    • 2 years 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
    • 2 years ago
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    I think it was Tukey

  72. help123please.
    • 2 years 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
    • 2 years ago
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    you just found it

  77. help123please.
    • 2 years 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
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    subtract that from the median 68

  81. help123please.
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    Hmm. Ok. WOW, lots of steps :/ LOL.

  82. jim_thompson5910
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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.
    • 2 years ago
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    o right sorry.

  88. jim_thompson5910
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    so that's your lowest threshold

  89. help123please.
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    i punched the wrong numbers into the calculator.

  90. jim_thompson5910
    • 2 years ago
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    basically that's the smallest the min can be

  91. help123please.
    • 2 years 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.
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    Ah,ok.

  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

  95. jim_thompson5910
    • 2 years ago
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    now compute median + 1.5*IQR

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

  97. help123please.
    • 2 years ago
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    ill brb

  98. jim_thompson5910
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    ok

  99. help123please.
    • 2 years ago
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    im back

  100. jim_thompson5910
    • 2 years ago
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    ok

  101. jim_thompson5910
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    did you find median + 1.5*IQR

  102. help123please.
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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
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    you already did that one

  106. help123please.
    • 2 years ago
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    its not the answer?

  107. jim_thompson5910
    • 2 years 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
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    yeah but the IQR is 6, not 9

  111. help123please.
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    right,sorry

  112. help123please.
    • 2 years 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
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    so there are no outliers

  121. jim_thompson5910
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    you're correct

  122. help123please.
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    nvm.

  123. help123please.
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    oh,okay.

  124. jim_thompson5910
    • 2 years 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
    • 2 years 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.
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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.
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    Okay. I think I understand a bit better- thanks!

  131. jim_thompson5910
    • 2 years ago
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    yw

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