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getzgetzu
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The average of w, x, y and z is m. S is the standard 21 Apr 2006, 03:47
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The average of w, x, y and z is m. S is the standard deviation. Is S>2?

1. w=3
2. m=8



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SimaQ
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The average of w, x, y and z is m. S is the standard 21 Apr 2006, 04:20
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Hmm....

1) statement is insufficient....

2) statement gives us that w+x+y+z=32.... I would choose B.....
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The average of w, x, y and z is m. S is the standard 21 Apr 2006, 04:45
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SimaQ
Hmm....

1) statement is insufficient....

2) statement gives us that w+x+y+z=32.... I would choose B.....
Show more


How do you get SD from Mean?
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SimaQ
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The average of w, x, y and z is m. S is the standard 21 Apr 2006, 04:59
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trivikram
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Hmm....

1) statement is insufficient....

2) statement gives us that w+x+y+z=32.... I would choose B.....

How do you get SD from Mean?
Show more


I dont :)

Standard deviation could be roughly defined as the spread between the numbers in the set..... so to get 32 from 4 numbers you have to have a big enough spread between the numbers...... 6 and 10 as the minimum...that is my reasoning
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The average of w, x, y and z is m. S is the standard 21 Apr 2006, 06:56
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I think ans = C

statement 1 gives nothing much.

From statement 2, we know W + X + Y + Z = 32. But we have no idea of SD.
if W=X = Y = Z, S < 2. With other cases, it can be > 2.

Combining these two, we know that s > 2
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The average of w, x, y and z is m. S is the standard 21 Apr 2006, 09:16
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I belive C is the answer...

for SD to be greater than 2 we need

(w-m)^2+(x-m)^2+(y-m)^2+(z-m)^2/4 > 2

=>(w-m)^2+(x-m)^2+(y-m)^2+(z-m)^2 > 8

now that means only one of w,x,y,z has to be more than 3 unit away from m. so for m=8 and w=3 it's going to be always >8.
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The average of w, x, y and z is m. S is the standard 21 Apr 2006, 10:42
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chiragr
I belive C is the answer...

for SD to be greater than 2 we need

(w-m)^2+(x-m)^2+(y-m)^2+(z-m)^2/4 > 2

=>(w-m)^2+(x-m)^2+(y-m)^2+(z-m)^2 > 8

now that means only one of w,x,y,z has to be more than 3 unit away from m. so for m=8 and w=3 it's going to be always >8.
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Agree with chiragr.

Given m = 8 and w =3

SD^2 = [(8-3)^2+(8-x)^2+(8-y)^2+(8-z)^2]/4
= (25+K)/4 --> K is some +ve value
SD = SQRT (25/4+K/4)
SQRT(25/4) = SQRT(6.25) = 2.5
Hence SD = 2.5+K' > 2
C it is!
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The average of w, x, y and z is m. S is the standard 21 Apr 2006, 20:31
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SimaQ
trivikram
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Hmm....

1) statement is insufficient....

2) statement gives us that w+x+y+z=32.... I would choose B.....

How do you get SD from Mean?

I dont :)

Standard deviation could be roughly defined as the spread between the numbers in the set..... so to get 32 from 4 numbers you have to have a big enough spread between the numbers...... 6 and 10 as the minimum...that is my reasoning
Show more


What if all the four numbers are same. Then SD would be 0 !!

IMO C.
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The average of w, x, y and z is m. S is the standard 10 Jun 2006, 00:06
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trivikram
SimaQ
Hmm....

1) statement is insufficient....

2) statement gives us that w+x+y+z=32.... I would choose B.....

How do you get SD from Mean?

I dont :)

Standard deviation could be roughly defined as the spread between the numbers in the set..... so to get 32 from 4 numbers you have to have a big enough spread between the numbers...... 6 and 10 as the minimum...that is my reasoning
Show more


I go for C in this one.
Why doesn't B stand?

m= 8 ..let's take example: w= 8.2 , x=8.1, y=7.9 , z= 7.8 --> m= 8
but SD = sqrt{[( 8.2-8)^2 + ( 8.1-8)^2 + (7.9 - 8) ^2 + ( 7.8-8)^2]/4 } is definitely smaller than 2

We can take another example to prove that SD can be bigger than 2
---> insuff.
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The average of w, x, y and z is m. S is the standard 10 Jun 2006, 05:45
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laxieqv
SimaQ
trivikram
SimaQ
Hmm....

1) statement is insufficient....

2) statement gives us that w+x+y+z=32.... I would choose B.....

How do you get SD from Mean?

I dont :)

Standard deviation could be roughly defined as the spread between the numbers in the set..... so to get 32 from 4 numbers you have to have a big enough spread between the numbers...... 6 and 10 as the minimum...that is my reasoning

I go for C in this one.
Why doesn't B stand?

m= 8 ..let's take example: w= 8.2 , x=8.1, y=7.9 , z= 7.8 --> m= 8
but SD = sqrt{[( 8.2-8)^2 + ( 8.1-8)^2 + (7.9 - 8) ^2 + ( 7.8-8)^2]/4 } is definitely smaller than 2

We can take another example to prove that SD can be bigger than 2
---> insuff.
Show more


B also does stand if w=x=y=z=8 !!
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The average of w, x, y and z is m. S is the standard 10 Jun 2006, 05:55
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but that case is a specific one. Since the question doesnt indicate anything about whether 4 numbers are equal or not, we have to consider both cases. Thus, B still doesn't stand.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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