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

Question

The average of w, x, y, and z is m. S is the deviation Is S>2?
1). w=3
2). m=8

goldeneagle94 · Answer

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{(8-3)^2 (8-9.6)^2 (8-9.6)^2 (8-9.6)^2}{4}}

Therefore, S > 2.

GMAT TIGER · Answer

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.

Neochronic · Answer

i first thought the ansewr to be E..

now its clear..

thanks for the approach..

vcbabu · Answer

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 .

lahoosaher · Answer

IMO C
Nice explanation goldeneagle94 .

tenaman10 · Answer

Any alternative ?? without calulating SD ..

ngoctraiden1905 · Answer

1) m, x, y, and z are unknown. NSF.
2) w, x, y and z are unknown. NSF.
Both, S = Sqrt  /4
S>2  /4 >4 (square each side)   > 16

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