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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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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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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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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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