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The average of w, x, y, and z is m. S is the deviation Is [#permalink]
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02 Jun 2009, 11:29
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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



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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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i first thought the ansewr to be E..
now its clear..
thanks for the approach..



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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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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










