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# Standard Deviation

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Senior Manager
Joined: 08 Nov 2008
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Standard Deviation  [#permalink]

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10 Mar 2009, 06:03
J is a collection of four odd integers whose range is 4. The standard deviation of J must be one of how many
numbers?
A. 3
B. 4
C. 5
D. 6
E. 7

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Manager
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Posts: 57
Re: Standard Deviation  [#permalink]

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10 Mar 2009, 09:30
Suppose the integers are 3, 5 and 7. The four integers can be

3,7,7,7
3,5,5,7
3,3,7,7
3,3,3,7
3,3,5,7
3,5,7,7

Each can have different SD. So 6
Manager
Joined: 05 Jan 2009
Posts: 73
Re: Standard Deviation  [#permalink]

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10 Mar 2009, 09:58
IMO 7(E). Need to take in account when all numbers are the same. SD=0.
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Re: Standard Deviation  [#permalink]

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10 Mar 2009, 10:02
B
1. Range of 4 means that at least two odd integers form boundaries of the range: X _ X, where X - odd integer, _ - a vacation.

2. Now let's add one more odd integer. There is 2 unique placements for third integer with respect to SD:
(X,X) _ X and X X X. X _ (X,X) has the same SD as the first option.

3. Let's add fourth odd integer:

(X,X,X) _ X
(X,X) X X
(X,X) _ (X,X)
X (X,X) X

Other options have the same SD, for example, (X,X) X X and X X (X,X).
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Re: Standard Deviation  [#permalink]

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10 Mar 2009, 10:06
pmal04 wrote:
IMO 7(E). Need to take in account when all numbers are the same. SD=0.

You forget about range=4. So all numbers cannot be the same.
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Re: Standard Deviation  [#permalink]

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10 Mar 2009, 10:08
kaushik04 wrote:
...
3,7,7,7
3,5,5,7
3,3,7,7
3,3,3,7
3,3,5,7
3,5,7,7
...

I've pointed out options with the same SD by red and blue colors.
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Manager
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Re: Standard Deviation  [#permalink]

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10 Mar 2009, 12:00
Hi..Thanks for pointing out. Can you tell me a easiest way to find the SD?

Is there a way to determine what the SD is by looking at the numbers if they have digits repeating?

walker wrote:
kaushik04 wrote:
...
3,7,7,7
3,5,5,7
3,3,7,7
3,3,3,7
3,3,5,7
3,5,7,7
...

I've pointed out options with the same SD by red and blue colors.
CEO
Joined: 17 Nov 2007
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Re: Standard Deviation  [#permalink]

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10 Mar 2009, 12:41
kaushik04 wrote:
Hi..Thanks for pointing out. Can you tell me a easiest way to find the SD?

Is there a way to determine what the SD is by looking at the numbers if they have digits repeating?

Here I didn't calculate SD to figure out that 3,3,3,7 and 3,7,7,7 have the same SD. It is just symmetry approach.
Anyway, in 95% cases it is helpful to think about SD as an average deviation of x from xav.
In other words SD = sqrt(sum(x^2-xav^2))/n ~ sum(|x-xav|)/n
For example, 3,3,3,7 has xav=16/4=4 and SD ~ (1+1+1+3)/4 ~ 1.5
3,7,7,7 has xav=24/4=6 and SD ~ (3+1+1+1)/4 ~ 1.5
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Re: Standard Deviation  [#permalink]

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10 Mar 2009, 13:29
walker wrote:
kaushik04 wrote:
Hi..Thanks for pointing out. Can you tell me a easiest way to find the SD?

Is there a way to determine what the SD is by looking at the numbers if they have digits repeating?

Here I didn't calculate SD to figure out that 3,3,3,7 and 3,7,7,7 have the same SD. It is just symmetry approach.
Anyway, in 95% cases it is helpful to think about SD as an average deviation of x from xav.
In other words SD = sqrt(sum(x^2-xav^2))/n ~ sum(|x-xav|)/n
For example, 3,3,3,7 has xav=16/4=4 and SD ~ (1+1+1+3)/4 ~ 1.5
3,7,7,7 has xav=24/4=6 and SD ~ (3+1+1+1)/4 ~ 1.5

Just to add to the above discussion.

if you calculate (element - mean) and sum them up for each of the above group in which SD is same, it comes to 0 for each of them.

ex: for 3,3,5,7 -----> -1.5,-1.5, .5, 2.5

if we sum these up its 0. same is the case for other three in the group. But again the sum comes to 0 for each of the 4 group and we can not say that SD is same for all of them. we've to use some kind of symmetry to diffrentiate between those. In this case the symmetry between 3,3,5,7 and 3,5,7,7 is like 4,4,8,10 and 4,6,10,10.
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Re: Standard Deviation  [#permalink]

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10 Mar 2009, 13:53
1
neeshpal wrote:
if you calculate (element - mean) and sum them up for each of the above group in which SD is same, it comes to 0 for each of them.

ex: for 3,3,5,7 -----> -1.5,-1.5, .5, 2.5

if we sum these up its 0. same is the case for other three in the group. But again the sum comes to 0 for each of the 4 group and we can not say that SD is same for all of them.

the sum of (element - mean) is always 0 for any set because it is a mean point and it is one of its definition. Maybe I haven't understood you correctly but I used |element - mean|

neeshpal wrote:
we've to use some kind of symmetry to differentiate between those. In this case the symmetry between 3,3,5,7 and 3,5,7,7 is like 4,4,8,10 and 4,6,10,10.

Perhaps, it would be also useful to mention some properties of SD and translate symmetry to math expressions:

1. SD({x})=SD({x+k})
2. SD({x})=SD({-x})

In our case with 3,3,5,7 we can use (10-x) transformation and get 7,7,5,3 set (10-3,10-3,10-5,10-7)
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Manager
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Location: nj
Re: Standard Deviation  [#permalink]

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10 Mar 2009, 14:28
walker wrote:
neeshpal wrote:
if you calculate (element - mean) and sum them up for each of the above group in which SD is same, it comes to 0 for each of them.

ex: for 3,3,5,7 -----> -1.5,-1.5, .5, 2.5

if we sum these up its 0. same is the case for other three in the group. But again the sum comes to 0 for each of the 4 group and we can not say that SD is same for all of them.

the sum of (element - mean) is always 0 for any set because it is a mean point and it is one of its definition. Maybe I haven't understood you correctly but I used |element - mean|

neeshpal wrote:
we've to use some kind of symmetry to differentiate between those. In this case the symmetry between 3,3,5,7 and 3,5,7,7 is like 4,4,8,10 and 4,6,10,10.

Perhaps, it would be also useful to mention some properties of SD and translate symmetry to math expressions:

1. SD({x})=SD({x+k})
2. SD({x})=SD({-x})

In our case with 3,3,5,7 we can use (10-x) transformation and get 7,7,5,3 set (10-3,10-3,10-5,10-7)

I was a bit confused within myself when i wrote the above comment. I did not mean to point out anything.

Thanks for these SD props.
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Re: Standard Deviation  [#permalink]

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11 Mar 2009, 07:58
OA:B
OE -
(B) For the sake of simplicity, let the smallest number in J be 1. Since the range is 4, the largest number in J is 5.
You may recall that the standard deviation of a set of numbers is the square root of its variance. How many
different variances are possible?
{ 1, 1, 1, 5 } and { 1, 5, 5, 5 } will have the same variance. Why? Remember that the variance is the average
squared difference between each element and the mean of the set of numbers. The first set above has a mean of
2, whereas the mean of the second set is 4– in the first set, three elements are 1 unit from the mean and the
other element is 3 units from the mean. This is true of the second set, too. These two sets, then, have a variance
of 12/4.
Similarly, { 1, 1, 3, 5 } (mean 2.5) and { 1, 3, 5, 5 } (mean 3.5) will have the same variance: in each set, two
elements are 1.5 units from the mean, one element is 0.5 unit from the mean, and the remaining element is 2.5
units from the mean– these two sets have a variance of 11/4.
Finally, {1, 3, 3, 5} and {1, 1, 5, 5} have variances of 8/4 and 16/4 respectively.
So, 4 different variances and thus four different standard deviations are possible. Answer: B

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Re: Standard Deviation &nbs [#permalink] 11 Mar 2009, 07:58
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# Standard Deviation

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