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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] New post 24 Jan 2008, 15:08
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A set of data consists of the following 5 numbers: 0,2,4,6 and 8. Which 2 numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is closest to the standard deviation for of the 5 original numbers?

A) -1 and 9
B) 4 and 4
C) 3 and 5
D) 2 and 6
E) 0 and 8

Sorry, I mistook and wrote d instead of 4 in B.

Last edited by netcaesar on 25 Jan 2008, 08:36, edited 1 time in total.
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Re: PS: Standard Deviation [#permalink] New post 24 Jan 2008, 17:44
Average is 4. Each of the pairs has an average of 4 except for B. which is 4 and D.

Answer B
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Re: PS: Standard Deviation [#permalink] New post 24 Jan 2008, 20:11
eschn3am wrote:
Average is 4. Each of the pairs has an average of 4 except for B. which is 4 and D.

Answer B


??????? How can you deduce that (4+D)/2 isn't going to be 4?
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Re: PS: Standard Deviation [#permalink] New post 24 Jan 2008, 20:29
GMATBLACKBELT wrote:
eschn3am wrote:
Average is 4. Each of the pairs has an average of 4 except for B. which is 4 and D.

Answer B


??????? How can you deduce that (4+D)/2 isn't going to be 4?



I can't. It's a case of knowing that all the other ones are 4 for sure, so I'll put my money on the one that might be something else.

I agree that it's bizarre logic though. I'm thinking there must be something wrong with either the question, my logic or my understanding of the question because choosing B for being the lesser of 5 evils is ridiculous.
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Re: PS: Standard Deviation [#permalink] New post 25 Jan 2008, 00:32
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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.
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Re: PS: Standard Deviation [#permalink] New post 25 Jan 2008, 08:37
Sorry, I just corrected the data of this problem!!!
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Re: PS: Standard Deviation [#permalink] New post 29 Jan 2008, 14:43
I don´t understand very well why D is the right answer and not the other ones.

Besides, is there any other way to solve it without using the formula? With the intervals of 2, perhaps???
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Re: PS: Standard Deviation [#permalink] New post 29 Jan 2008, 14:52
netcaesar wrote:
A set of data consists of the following 5 numbers: 0,2,4,6 and 8. Which 2 numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is closest to the standard deviation for of the 5 original numbers?

A) -1 and 9
B) 4 and 4
C) 3 and 5
D) 2 and 6
E) 0 and 8

Sorry, I mistook and wrote d instead of 4 in B.


Average of 5 numbers = 4
The way I look at it is each of the choices will not affect the average (it remains 4 as the total will be 28 for 7 numbers), but the standard deviation will be least affected by choice B simply because (4-4)^2 will not affect the overall summation.

The answer should be (B)
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Re: PS: Standard Deviation [#permalink] New post 29 Jan 2008, 18:12
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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
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Re: PS: Standard Deviation [#permalink] New post 29 Jan 2008, 18:33
walker wrote:
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


walker, you're right -- I missed the division by N in the standard deviation calculation :)
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Re: PS: Standard Deviation [#permalink] New post 30 Jan 2008, 09:23
I thought in another way but I am not sure if it is true:

We know that the Av=4 in the group.

There is an interval of 2 units in the group: 0,2,4,6 and8.

So, can I conclude that the minimum difference in the St. Deviation will be with another 2 elements 2 points far from the average (that is,2 and 6)?
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Re: PS: Standard Deviation [#permalink] New post 30 Jan 2008, 11:59
Expert's post
netcaesar wrote:
I thought in another way but I am not sure if it is true:

We know that the Av=4 in the group.

There is an interval of 2 units in the group: 0,2,4,6 and8.

So, can I conclude that the minimum difference in the St. Deviation will be with another 2 elements 2 points far from the average (that is,2 and 6)?


I think It is too risky in general case.
"Ideally, we have to add two numbers for which(\sum{(y-y_{av})^2})_{av}=8" - is one of the right ways.

And thanks for your PS question. :)
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Re: PS: Standard Deviation   [#permalink] 30 Jan 2008, 11:59
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