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The average of w, x, y, and z is m. S is the deviation Is
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02 Jun 2009, 12:29
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The average of w, x, y, and z is m. S is the deviation Is S>2? 1). w=3 2). m=8 == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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Joined: 08 Feb 2009
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Answer  C
Clearly, we need both the statments.
So, combining
w = 3 & m = 8
\(\frac{w+x+y+z}{4} = 8\)
\(\Rightarrow \frac{x+y+z}{3} = \frac{29}{3} = 9.6 (approx)\)
Standard deviation will be lowest, when x,y,z are equal to 9.6 OR sum of x,y,z is equal to 29.
In such a case, Standard Deviation = \(S = \sqrt{\frac{(83)^2 (89.6)^2 (89.6)^2 (89.6)^2}{4}}\)
Therefore, S > 2.



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vcbabu wrote: The average of w, x, y, and z is m. S is the deviation Is S>2? 1). w=3 2). m=8 1) m, x, y, and z are unknown. NSF. 2) w, x, y and z are unknown. NSF. C. sqrt (wm)/4 = sqrt(83)/ = sqrt(6.25), which is > 2. No matter the value of x, y and z, SD is > 2.
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Senior Manager
Joined: 15 Jan 2008
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i first thought the ansewr to be E..
now its clear..
thanks for the approach..



Manager
Joined: 04 Sep 2006
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1) m, x, y, and z are unknown. NSF. 2) w, x, y and z are unknown. NSF.
C. sqrt (wm)/4 = sqrt(83)/ = sqrt(6.25), which is > 2. No matter the value of x, y and z, SD is > 2.
how did u decide .



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IMO C Nice explanation goldeneagle94 .
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Senior Manager
Joined: 08 Nov 2008
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Any alternative ?? without calulating SD .. goldeneagle94 wrote: Answer  C
Clearly, we need both the statments.
So, combining
w = 3 & m = 8
\(\frac{w+x+y+z}{4} = 8\)
\(\Rightarrow \frac{x+y+z}{3} = \frac{29}{3} = 9.6 (approx)\)
Standard deviation will be lowest, when x,y,z are equal to 9.6 OR sum of x,y,z is equal to 29.
In such a case, Standard Deviation = \(S = \sqrt{\frac{(83)^2 (89.6)^2 (89.6)^2 (89.6)^2}{4}}\)
Therefore, S > 2.
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1) m, x, y, and z are unknown. NSF. 2) w, x, y and z are unknown. NSF. Both, S = Sqrt [(xm)^2 + (ym)^2 +(zm)^2 +(wm)^2]/4 S>2 [(xm)^2 + (ym)^2 +(zm)^2 +(wm)^2]/4 >4 (square each side) [(xm)^2 + (ym)^2 +(zm)^2 +(38)^2] > 16 [(xm)^2 + (ym)^2 +(zm)^2 + 25 >16 Obviously Hence C == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.










