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laxieqv
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A list w,x,y,z, standard deviation d = sqrt( ((w-m)^2 + 02 Mar 2006, 12:06
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A list w,x,y,z, standard deviation d = sqrt( ((w-m)^2 + (x-m)^2 + (y-m)^2 + (z-m)^2 )/4) ), is d>2?
(1) w = 3
(2) m = 8



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A list w,x,y,z, standard deviation d = sqrt( ((w-m)^2 + 02 Mar 2006, 12:22
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Halllo,
think that both A) and B) are by themselves insufficent cause we have no info for the other values.
Both together are sufficient to answer the Q. This SD is MIN when x,y,z are equal to the average m. Then SD=5/2. No matter what other values these x,y,z can take the SD will be >5/2
So C)
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A list w,x,y,z, standard deviation d = sqrt( ((w-m)^2 + 02 Mar 2006, 21:49
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E. We don't know if w is the smalles or largest element in the list.
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A list w,x,y,z, standard deviation d = sqrt( ((w-m)^2 + 02 Mar 2006, 22:06
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laxieqv
A list w,x,y,z, standard deviation, d = sqrt(((w-m)^2 + (x-m)^2 + (y-m)^2 + (z-m)^2 )/4) ), is d>2?
(1) w = 3
(2) m = 8
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C.
yes d>2 because no no matter what other values are d = sqrt [{(w-m)^2 + (x-m)^2 + (y-m)^2 + (z-m)^2)/4)] = at least sqrt(25/4)=at least 5/2 which is greater than 2.

it is indeed a good question..
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A list w,x,y,z, standard deviation d = sqrt( ((w-m)^2 + 02 Mar 2006, 22:13
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Professor
laxieqv
A list w,x,y,z, standard deviation, d = sqrt(((w-m)^2 + (x-m)^2 + (y-m)^2 + (z-m)^2 )/4) ), is d>2?
(1) w = 3
(2) m = 8

C.
yes d>2 because no no matter what other values are d = sqrt [{(w-m)^2 + (x-m)^2 + (y-m)^2 + (z-m)^2)/4)] = at least sqrt(25/4)=at least 5/2 which is greater than 2.

it is indeed a good question..
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could u explain further please. how did u get sqrt(25/4)
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A list w,x,y,z, standard deviation d = sqrt( ((w-m)^2 + 02 Mar 2006, 22:46
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trublu
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laxieqv
A list w,x,y,z, standard deviation, d = sqrt(((w-m)^2 + (x-m)^2 + (y-m)^2 + (z-m)^2 )/4) ), is d>2?
(1) w = 3
(2) m = 8
C.
yes d>2 because no no matter what other values are d = sqrt [{(w-m)^2 + (x-m)^2 + (y-m)^2 + (z-m)^2)/4)] = at least sqrt(25/4)=at least 5/2 which is greater than 2.
it is indeed a good question..
could u explain further please. how did u get sqrt(25/4)
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from i and ii, w = 3 and m = 8, where m is mean and we only know the values of m and w. we donot know the values of other variables x, y and z.

d = sqrt [{(w-m)^2 + (x-m)^2 + (y-m)^2 + (z-m)^2)/4)]
d = sqrt [{(3-8)^2 + (x-m)^2 + (y-m)^2 + (z-m)^2)/4)]

here you can suppose any values for x, y and z. no mathher what the values of x, y and z, the values of [(x-m)^2 + (y-m)^2 + (z-m)^2] cannot be less than 0. so lets suppose the value of [(x-m)^2 + (y-m)^2 + (z-m)^2] = 0.
d = sqrt [{(-5)^2 + 0 )/4)]
d = sqrt (25/4)
d = 5/2 because SD cannot be in -ve.
d = 5/2>2.
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A list w,x,y,z, standard deviation d = sqrt( ((w-m)^2 + 04 Mar 2006, 08:49
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Professor
laxieqv
A list w,x,y,z, standard deviation, d = sqrt(((w-m)^2 + (x-m)^2 + (y-m)^2 + (z-m)^2 )/4) ), is d>2?
(1) w = 3
(2) m = 8

C.
yes d>2 because no no matter what other values are d = sqrt [{(w-m)^2 + (x-m)^2 + (y-m)^2 + (z-m)^2)/4)] = at least sqrt(25/4)=at least 5/2 which is greater than 2.

it is indeed a good question..
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this is indeed a good explanation, buddy~! :wink:

OA is C.
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A list w,x,y,z, standard deviation d = sqrt( ((w-m)^2 + 06 Jun 2007, 23:41
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good question and good to know
thanks all!
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shahrukh
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A list w,x,y,z, standard deviation d = sqrt( ((w-m)^2 + 06 Jun 2007, 23:55
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wow. this one flew right above me
looks damn difficult but is pretty darn simple
great question



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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