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anonymous

  • one year ago

54,55,59,61,61,62,68,70,72 how do i find the irst quartile, median, third quartile, and maximum of the data set?

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  1. LynFran
    • one year ago
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    the first quartile can be found using this formula \[\frac{ 1 }{ 4 }(n+1)\]

  2. LynFran
    • one year ago
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    n=9

  3. anonymous
    • one year ago
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    9/4?

  4. LynFran
    • one year ago
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    since there 9 #s in the set

  5. anonymous
    • one year ago
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    Okay

  6. LynFran
    • one year ago
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    \[\frac{ 1 }{ 4 }(9+1)\]

  7. anonymous
    • one year ago
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    1/4(9+1

  8. anonymous
    • one year ago
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    1 can't be divided by 4 can it?

  9. anonymous
    • one year ago
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    Do i move the 9+1 over the 4.

  10. LynFran
    • one year ago
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    u will get \[\frac{ 1 }{ 4 }(10)\]\[2.5th value\] the average 2nd and 3rd values are 55 and 59 so \[\frac{ 55+59 }{ 2 }=?\]

  11. anonymous
    • one year ago
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    Lagging

  12. LynFran
    • one year ago
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    55+59/2=57 now 57 is ur 1st quartile value

  13. anonymous
    • one year ago
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    Okay. what is the 2.5th value?

  14. LynFran
    • one year ago
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    thats the average where u will find ur two numbers for the fisrt quartile

  15. LynFran
    • one year ago
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    so in this set of numbers the first quartile is between the the 2nd and 3rd #

  16. anonymous
    • one year ago
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    Okay

  17. LynFran
    • one year ago
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    so we add the 2nd and 3rd # and then divide the # by 2

  18. anonymous
    • one year ago
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    okay would that be 54 and 55?

  19. LynFran
    • one year ago
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    no thats the 1st and 2nd # ... u need the 2nd and 3rd #

  20. anonymous
    • one year ago
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    so 55 and 59?

  21. LynFran
    • one year ago
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    yes now add them

  22. anonymous
    • one year ago
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    100 +14 114

  23. LynFran
    • one year ago
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    59+55=114 now we divide by 2

  24. LynFran
    • one year ago
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    114/2=?

  25. anonymous
    • one year ago
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    57

  26. LynFran
    • one year ago
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    good now 57 is ur 1st quartile

  27. anonymous
    • one year ago
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    Alright.

  28. LynFran
    • one year ago
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    to find the median we located the middle value in the set....what # is in the middle?

  29. anonymous
    • one year ago
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    61?

  30. LynFran
    • one year ago
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    correct

  31. LynFran
    • one year ago
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    to find the 3rd quartile the formula is \[\frac{ 3 }{ 4 }(n+1)th . value \]

  32. LynFran
    • one year ago
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    \[\frac{ 3 }{ 4 }(9+1)th.value\]

  33. anonymous
    • one year ago
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    What do i do first add the 9+1 will give me 10 but what do i do next?

  34. LynFran
    • one year ago
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    \[\frac{ 3 }{4 }(10)th.value\]\[\frac{ 30 }{ 4 }th. value\]7.5 th. value so the 3rd quartile is between the 7th and 8th numbers

  35. anonymous
    • one year ago
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    Oh so we time it by 3 3*10 = 30 /4

  36. anonymous
    • one year ago
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    7th and 8th what does that mean?

  37. LynFran
    • one year ago
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    we locate the 7th and 8th #s in the data/set

  38. LynFran
    • one year ago
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    whats the 7th and 8 number frm the set

  39. anonymous
    • one year ago
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    Okay.

  40. anonymous
    • one year ago
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    68 and 70

  41. LynFran
    • one year ago
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    good now add them and divide by 2

  42. anonymous
    • one year ago
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    138/2 = 69

  43. LynFran
    • one year ago
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    good thats ur 3rd quartile

  44. anonymous
    • one year ago
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    So we should be done.

  45. LynFran
    • one year ago
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    whats the maximum?

  46. anonymous
    • one year ago
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    Not sure. ^ ^;

  47. LynFran
    • one year ago
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    72..

  48. anonymous
    • one year ago
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    DX Ahh.

  49. anonymous
    • one year ago
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    Okay thanks.

  50. LynFran
    • one year ago
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    here more info on quartiles with a similar example http://www.mathsteacher.com.au/year9/ch17_statistics/06_quartiles/quartiles.htm

  51. LynFran
    • one year ago
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    welcome

  52. anonymous
    • one year ago
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    Thanks for the link too.

  53. LynFran
    • one year ago
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    the 2nd quatile \[\frac{ 2 }{ 4 }(10)th.value \]= 5th. value so the 5th value is 61 and yes 61 is the 2nd quartile

  54. anonymous
    • one year ago
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    Alright thanks.

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