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D01-18

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Bunuel
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D01-18 [#permalink] New post   16 Sep 2014, 00:12
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If data set \(S\) consists of more than two elements, what is the standard deviation of \(S\)?


(1) The mean and the median of the data set are positive.

(2) The difference between any two elements of the data set is equal.
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Bunuel
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Re D01-18 [#permalink] New post   16 Sep 2014, 00:12
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Official Solution:


If data set \(S\) consists of more than two elements, what is the standard deviation of \(S\)?

(1) The mean and the median of the data set are positive. The above is clearly insufficient: a positive mean and median does not provide any insight into the standard deviation.

(2) The difference between any two elements of the data set is equal.

Since the difference between ANY two elements of the data set is equal and the data set contains more than two elements, it can be concluded that all elements of the data set are identical. The standard deviation of a data set consisting of identical elements is 0. Sufficient.


Answer: B
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Re: D01-18 [#permalink] New post   02 Dec 2014, 12:20
Bunuel wrote:
Official Solution:


Statement 1: If the mean and median of the set is positive, the standard deviation could be any. The set could have elements {1, 1, 1} or {1, 2, 3} or {10, 20, 30, 40, 50}. In each case, the standard deviation isn’t the same. So NSF.

Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


Answer: B


Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


I'm confused why they have to be the same elements because the number of elements is greater than 2.. If the difference is equal, can't it just be {1,3,5..} or {1,5,9..} which means SD can be anything..

Also, can you explain difference between elements and numbers in this case? This may be adding to my confusion.
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Re: D01-18 [#permalink] New post   03 Dec 2014, 04:06
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codeblue wrote:
Bunuel wrote:
Official Solution:


Statement 1: If the mean and median of the set is positive, the standard deviation could be any. The set could have elements {1, 1, 1} or {1, 2, 3} or {10, 20, 30, 40, 50}. In each case, the standard deviation isn’t the same. So NSF.

Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


Answer: B


Statement 2: If difference between any elements of the set is equal, then the set has to have same elements because the number of elements is greater than 2. So standard deviation is 0. Sufficient.


I'm confused why they have to be the same elements because the number of elements is greater than 2.. If the difference is equal, can't it just be {1,3,5..} or {1,5,9..} which means SD can be anything..

Also, can you explain difference between elements and numbers in this case? This may be adding to my confusion.


Second statement says that the difference between ANY two elements of the set is equal. If the set does not have all the elements equal, for example, if the set is {1, 3, 5}, then the difference between ANY two elements of the set won't be equal: 3-1=2 but 5-1=4. Hence the set must have same elements.

As for your other question: element of a set and number of a set are the same thing - member of a set.
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Re: D01-18 [#permalink] New post   19 Oct 2020, 23:49
I think this the explanation isn't clear enough, please elaborate. In the 2nd statement
I assumed number of elements to be 3 and elements as
a,a+d,a+2d
Using std deviation formula we get=d√(2/3)
But d remains ambiguous.
Is it ok?
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Re: D01-18 [#permalink] New post   19 Oct 2020, 23:54
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ironsid wrote:
I think this the explanation isn't clear enough, please elaborate. In the 2nd statement
I assumed number of elements to be 3 and elements as
a,a+d,a+2d
Using std deviation formula we get=d√(2/3)
But d remains ambiguous.
Is it ok?


If the list is {a, a + d, a + 2d}, the difference between ANY two elements of the list is NOT equal. For example, (a + 2d) - a = 2d but (a + 2d) - (a + d) = d. For the difference between ANY two elements of the list to be equal, the list has to have same elements because the number of elements is greater than 2.
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Re: D01-18 [#permalink] New post   01 Mar 2023, 03:34
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: D01-18 [#permalink]
01 Mar 2023, 03:34
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