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A set of data consists of the following 5 numbers: 0,2,4,6, [#permalink]
 15 Jul 2008, 09:41
50. A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?
A). -1 and 9
B). 4 and 4
C). 3 and 5
D). 2 and 6
E). 0 and 8
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Re: standard deviation [#permalink]
15 Jul 2008, 09:57
may be 3 and 5
snce 2 and 6 gives exactly same value as previous SD
what is OA?
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Re: standard deviation [#permalink]
15 Jul 2008, 10:28
What I've done.
Steps:
1- Avg = 4
2- Std Dev = (4+2+0+2+4)/5 = 12/5
3- New Std Dev = (12/5)*7 = 16,8
4- New Numbers Distance from Avg = (16,8 - 12)/2 = 2,4
5- New Numbers = 4 +2,4 or 4 -2,4
The closest is 2 and 6, therefore IMO D.
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Re: standard deviation [#permalink]
15 Jul 2008, 10:56
SD of original 5 = sqrt(8)
A- SD = sqrt(12.xx)
B- SD = sqrt(40/7) = sqrt(5.7)
C- SD = sqrt(6)
D- SD = sqrt(48/7) = sqrt(7)
E- SD = sqrt(72/7) = sqrt(10.3)
D is closest I hope this one is not a GMAT/GMATPrep material as I see no short cut for it.
***EDIT****
WHAT IS THE SOURCE OF THIS QUESTION? I hope not GPrep!
Last edited by mbawaters on 15 Jul 2008, 11:33, edited 2 times in total.
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Re: standard deviation [#permalink]
15 Jul 2008, 11:10
Compute standard deviation s of the original 5 :
s^2 = \frac{4^2+2^2+0+2^2+4^2}{5} = 8
You want to add two new numbers and want s^2 to stay close of 8.
New s^2 will be : s^2 = \frac{8*5+2*(X-4)^2}{7} (since 4 remains the average of the new set whichever choice you pick)
So if we want it to remain close to 8 we want to be close to \frac{8*5+2*(X-4)^2}{7} = 8
This leads to (X-4)^2 = 8 and thus X = 4 + 2 sqrt(2) or X = 4 - 2 sqrt(2)
sqrt(2) is close to 1.4, so we want X as close to 1.2 or 6.8 ==> answer (D)
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Re: standard deviation [#permalink]
15 Jul 2008, 12:49
i say B.
rem formula of S.D
from the formula, higher S.D means that higher is the spread. so by adding 4 and 4, the spread does not increase(since the mean is 4), at the same time, as we are increasing the no. of samples(from 5 to 7), the denominator in the formula (N) increases, thereby decreasing the S.D.
its an old problem....
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Re: standard deviation [#permalink]
15 Jul 2008, 12:55
arjtryarjtry wrote: i say B.
rem formula of S.D
from the formula, higher S.D means that higher is the spread. so by adding 4 and 4, the spread does not increase(since the mean is 4), at the same time, as we are increasing the no. of samples(from 5 to 7), the denominator in the formula (N) increases, thereby decreasing the S.D.
its an old problem.... Yeah but you'r now dividing it by 7 instead of 5...
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Re: standard deviation [#permalink]
16 Jul 2008, 05:35
what's d oa? without calculation i also picked b. but after looking at the other solutions, d seemed to be the correct answer
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Re: standard deviation [#permalink]
16 Jul 2008, 08:43
My fast approach to SD. (10-20sec) Think of SD as average deviation of data from a mean. the closer deviations to SD we add the less we change SD. 0,2,4,6,8 has mean of 4 and average deviation of 2. Therefore, 6 and 2 add the least changes in SD: 2=(6-4) and 2=(4-2). So, D
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Re: standard deviation [#permalink]
16 Jul 2008, 08:52
walker wrote: My fast approach to SD. (10-20sec)
Think of SD as average deviation of data from a mean.
the closer deviations to SD we add the less we change SD. 0,2,4,6,8 has mean of 4 and average deviation of 2. Therefore, 6 and 2 add the least changes in SD: 2=(6-4) and 2=(4-2). So, D walker can you ellaborate? knowing that 'SD is a deviation of data from a mean' how did you reject '4 and 4' when mean is 4? Thanks
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Re: standard deviation [#permalink]
16 Jul 2008, 09:58
walker wrote: My fast approach to SD. (10-20sec)
Think of SD as average deviation of data from a mean.
the closer deviations to SD we add the less we change SD. 0,2,4,6,8 has mean of 4 and average deviation of 2. Therefore, 6 and 2 add the least changes in SD: 2=(6-4) and 2=(4-2). So, D Walker how did u calculate the avg SD to be 2 mbawaters wrote: SD of original 5 = sqrt(8)
A- SD = sqrt(12.xx)
B- SD = sqrt(40/7) = sqrt(5.7)
C- SD = sqrt(6)
D- SD = sqrt(48/7) = sqrt(7)
E- SD = sqrt(72/7) = sqrt(10.3)
D is closest I hope this one is not a GMAT/GMATPrep material as I see no short cut for it.
***EDIT****
WHAT IS THE SOURCE OF THIS QUESTION? I hope not GPrep! I picked up this Q from a compilation of tough Qs that have appeared on GMAT ..There is a document that another friend of mine had given me.M not sure whether these really appeared on GMAT but i just do them when i don't feel like doing verbal ..  I am not sure whether i can attach the file on this forum ... The explanation given there is something like this SD =Sqrt (sum (X-x)^2 /N) Since N is changing from 5 to 7 . Value of sum(X-X)^2 should change from 40 (current) to 48. So that SD remains same. Here y should it change to 48 and not 56???
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Re: standard deviation [#permalink]
16 Jul 2008, 12:59
mbawaters wrote: walker can you ellaborate? knowing that 'SD is a deviation of data from a mean' how did you reject '4 and 4' when mean is 4?
Thanks MamtaKrishnia wrote: Walker how did u calculate the avg SD to be 2
I would say that this problem cannot be solved in a such way, because really deviation is 3 (2 was my mistake) and we have two close answer: D and E. see here for another approach: 7-t58976
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Re: standard deviation [#permalink]
16 Jul 2008, 15:58
MamtaKrishna,
It should change to 56. But answer choice D gives the change to 48 that is closest to 56.
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Re: standard deviation [#permalink]
17 Jul 2008, 05:33
scthakur wrote: MamtaKrishna,
It should change to 56. But answer choice D gives the change to 48 that is closest to 56. Thanks scthakur, This makes sense to me. Walker , In the other link you have posted one plausible explanation for this Q ... A. D(SET 5, -1, 9) = 3.5857 B. D(SET 5, 4, 4) = 2.3905 C. D(SET 5, 3, 5) = 2.4495 D. D(SET 5, 2, 6) = 2.6186 ---D(SET 5) = 2.8284 E. D(SET 5, 0, 8) = 3.2071 I didnt get this explanation
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Re: standard deviation [#permalink]
17 Jul 2008, 07:42
MamtaKrishnia wrote: Walker ,
In the other link you have posted one plausible explanation for this Q ...
A. D(SET 5, -1, 9) = 3.5857
B. D(SET 5, 4, 4) = 2.3905
C. D(SET 5, 3, 5) = 2.4495
D. D(SET 5, 2, 6) = 2.6186
---D(SET 5) = 2.8284
E. D(SET 5, 0, 8) = 3.2071
I didnt get this explanation His explanation was not that one (this is just a numerical approach with a computer: you cannot do it during the GMAT), but that one: walker wrote: D
x_{av}=(0+2+4+6+8)/5=4
(\sum{(x-x_{av})^2})_{av}=(16+4+0+4+16)/5=8
for all variants y_{av}=4
Ideally, we have to add two numbers for which(\sum{(y-y_{av})^2})_{av}=8
D has the closest value - 4. I did the same above and developped it a bit: Oski wrote: Compute standard deviation s of the original 5 :
s^2 = \frac{4^2+2^2+0+2^2+4^2}{5} = 8
You want to add two new numbers and want s^2 to stay close of 8.
New s^2 will be : s^2 = \frac{8*5+2*(X-4)^2}{7} (since 4 remains the average of the new set whichever choice you pick)
So if we want it to remain close to 8 we want to be close to \frac{8*5+2*(X-4)^2}{7} = 8
This leads to (X-4)^2 = 8 and thus X = 4 + 2 sqrt(2) or X = 4 - 2 sqrt(2)
sqrt(2) is close to 1.4, so we want X as close to 1.2 or 6.8 ==> answer (D)
Last edited by Oski on 17 Jul 2008, 07:43, edited 1 time in total.
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Re: standard deviation [#permalink]
17 Jul 2008, 07:43
MamtaKrishnia wrote:
Walker ,
In the other link you have posted one plausible explanation for this Q ...
A. D(SET 5, -1, 9) = 3.5857
B. D(SET 5, 4, 4) = 2.3905
C. D(SET 5, 3, 5) = 2.4495
D. D(SET 5, 2, 6) = 2.6186
---D(SET 5) = 2.8284
E. D(SET 5, 0, 8) = 3.2071
I didnt get this explanation
It is not my explanation, it is rather Excel proof that D is a right answer. My explanation here in another post of the thread: p424757#p424757
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Re: standard deviation [#permalink]
17 Jul 2008, 07:44
Thanks, Oski You were a little bit faster 
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Re: standard deviation [#permalink]
19 Jul 2008, 07:13
Oski, Your explanation makes things a lot clear ... Well m sure now ill be able to apply this logic if required on the actual GMAT .. 
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Re: standard deviation
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19 Jul 2008, 07:13
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