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# The average of w, x, y, and z is m. S is the deviation Is

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Manager
Joined: 04 Sep 2006
Posts: 113
The average of w, x, y, and z is m. S is the deviation Is  [#permalink]

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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

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Manager
Joined: 08 Feb 2009
Posts: 139
Schools: Anderson

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02 Jun 2009, 13:14

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 = $$S = \sqrt{\frac{(8-3)^2 (8-9.6)^2 (8-9.6)^2 (8-9.6)^2}{4}}$$

Therefore, S > 2.
SVP
Joined: 29 Aug 2007
Posts: 2419

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02 Jun 2009, 14:13
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 (w-m)/4 = sqrt(8-3)/ = 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
Posts: 260

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03 Jun 2009, 23:57
i first thought the ansewr to be E..

now its clear..

thanks for the approach..
Manager
Joined: 04 Sep 2006
Posts: 113

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04 Jun 2009, 10:56
1) m, x, y, and z are unknown. NSF.
2) w, x, y and z are unknown. NSF.

C. sqrt (w-m)/4 = sqrt(8-3)/ = sqrt(6.25), which is > 2. No matter the value of x, y and z, SD is > 2.

how did u decide .
Senior Manager
Joined: 16 Jan 2009
Posts: 338
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

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04 Jun 2009, 15:40
IMO C
Nice explanation goldeneagle94 .
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Lahoosaher

Senior Manager
Joined: 08 Nov 2008
Posts: 258

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06 Jun 2009, 06:35
Any alternative ?? without calulating SD ..
goldeneagle94 wrote:

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 = $$S = \sqrt{\frac{(8-3)^2 (8-9.6)^2 (8-9.6)^2 (8-9.6)^2}{4}}$$

Therefore, S > 2.

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"CEO in making"

Manager
Joined: 11 Aug 2008
Posts: 132

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08 Jun 2009, 23:46
1) m, x, y, and z are unknown. NSF.
2) w, x, y and z are unknown. NSF.
Both, S = Sqrt [(x-m)^2 + (y-m)^2 +(z-m)^2 +(w-m)^2]/4
S>2 [(x-m)^2 + (y-m)^2 +(z-m)^2 +(w-m)^2]/4 >4 (square each side) [(x-m)^2 + (y-m)^2 +(z-m)^2 +(3-8)^2] > 16 [(x-m)^2 + (y-m)^2 +(z-m)^2 + 25 >16 Obviously
Hence C

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Re: sd &nbs [#permalink] 08 Jun 2009, 23:46
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