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# A set of data consists of the following 5 numbers: 0,2,4,6,

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A set of data consists of the following 5 numbers: 0,2,4,6, [#permalink]

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15 Jul 2008, 09:41
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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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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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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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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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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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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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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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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16 Jul 2008, 08:43
1
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Expert's post
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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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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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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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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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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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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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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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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17 Jul 2008, 07:44
Thanks, Oski

You were a little bit faster
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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 ..
Re: standard deviation   [#permalink] 19 Jul 2008, 07:13
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