anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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LynFran
  • LynFran
the first quartile can be found using this formula \[\frac{ 1 }{ 4 }(n+1)\]
LynFran
  • LynFran
n=9
anonymous
  • anonymous
9/4?

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LynFran
  • LynFran
since there 9 #s in the set
anonymous
  • anonymous
Okay
LynFran
  • LynFran
\[\frac{ 1 }{ 4 }(9+1)\]
anonymous
  • anonymous
1/4(9+1
anonymous
  • anonymous
1 can't be divided by 4 can it?
anonymous
  • anonymous
Do i move the 9+1 over the 4.
LynFran
  • LynFran
u will get \[\frac{ 1 }{ 4 }(10)\]\[2.5th value\] the average 2nd and 3rd values are 55 and 59 so \[\frac{ 55+59 }{ 2 }=?\]
anonymous
  • anonymous
Lagging
LynFran
  • LynFran
55+59/2=57 now 57 is ur 1st quartile value
anonymous
  • anonymous
Okay. what is the 2.5th value?
LynFran
  • LynFran
thats the average where u will find ur two numbers for the fisrt quartile
LynFran
  • LynFran
so in this set of numbers the first quartile is between the the 2nd and 3rd #
anonymous
  • anonymous
Okay
LynFran
  • LynFran
so we add the 2nd and 3rd # and then divide the # by 2
anonymous
  • anonymous
okay would that be 54 and 55?
LynFran
  • LynFran
no thats the 1st and 2nd # ... u need the 2nd and 3rd #
anonymous
  • anonymous
so 55 and 59?
LynFran
  • LynFran
yes now add them
anonymous
  • anonymous
100 +14 114
LynFran
  • LynFran
59+55=114 now we divide by 2
LynFran
  • LynFran
114/2=?
anonymous
  • anonymous
57
LynFran
  • LynFran
good now 57 is ur 1st quartile
anonymous
  • anonymous
Alright.
LynFran
  • LynFran
to find the median we located the middle value in the set....what # is in the middle?
anonymous
  • anonymous
61?
LynFran
  • LynFran
correct
LynFran
  • LynFran
to find the 3rd quartile the formula is \[\frac{ 3 }{ 4 }(n+1)th . value \]
LynFran
  • LynFran
\[\frac{ 3 }{ 4 }(9+1)th.value\]
anonymous
  • anonymous
What do i do first add the 9+1 will give me 10 but what do i do next?
LynFran
  • LynFran
\[\frac{ 3 }{4 }(10)th.value\]\[\frac{ 30 }{ 4 }th. value\]7.5 th. value so the 3rd quartile is between the 7th and 8th numbers
anonymous
  • anonymous
Oh so we time it by 3 3*10 = 30 /4
anonymous
  • anonymous
7th and 8th what does that mean?
LynFran
  • LynFran
we locate the 7th and 8th #s in the data/set
LynFran
  • LynFran
whats the 7th and 8 number frm the set
anonymous
  • anonymous
Okay.
anonymous
  • anonymous
68 and 70
LynFran
  • LynFran
good now add them and divide by 2
anonymous
  • anonymous
138/2 = 69
LynFran
  • LynFran
good thats ur 3rd quartile
anonymous
  • anonymous
So we should be done.
LynFran
  • LynFran
whats the maximum?
anonymous
  • anonymous
Not sure. ^ ^;
LynFran
  • LynFran
72..
anonymous
  • anonymous
DX Ahh.
anonymous
  • anonymous
Okay thanks.
LynFran
  • LynFran
here more info on quartiles with a similar example http://www.mathsteacher.com.au/year9/ch17_statistics/06_quartiles/quartiles.htm
LynFran
  • LynFran
welcome
anonymous
  • anonymous
Thanks for the link too.
LynFran
  • LynFran
the 2nd quatile \[\frac{ 2 }{ 4 }(10)th.value \]= 5th. value so the 5th value is 61 and yes 61 is the 2nd quartile
anonymous
  • anonymous
Alright thanks.

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