Set M contains seven integers and set N contains three values chosen f

jabhatta2 it would be helpful if you could articulate why you expected what you expected.

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Let M = {1,2,3,4,5,6,7}
Mean(M) = 4
Variance = [(4-1)^2 + (4-2)^2 + (4-3)^2 + (4-4)^2 + (4-5)^2 + (4-6)^2 + (4-7)^2]/7 = 4
SD(M) = Root(Variance) = 2

Set N contains 3 values, all taken from Set M
Is SD(N) > 2? Or Is Variance(N) > 4?

(1) Set N contains the median of set M.

If N = {1,4,7}:
Mean = 4
Variance = [(4-1)^2 + (4-4)^2 + (4-7)^2]/3 = 6
SD(N) is greater than SD(M)

If N = {2,4,6}
Mean = 4
Variance = [(4-2)^2 + (4-4)^2 + (4-6)^2]/3 = 8/3
SD(N) is less than SD(M)

Not sufficient


(2) The range of set M and set N are equal.

Range of N = 6 which means that N has to contain elements 1 and 7
Which leaves us to the last element
Let us consider the 3 possible cases of different SD

Case 1: 3rd element = 2, Mean=3.33
Variance = [(3.33-1)^2 + (3.33-2)^2 + (3.33-7)^2]/3 = 6.88

Case 2: 3rd element = 3, Mean=3.66
Variance = [(3.66-1)^2 + (3.66-3)^2 + (3.66-7)^2]/3 = 6.22

Case 3: 3rd element= 4, Mean=4
Variance = [(4-1)^2 + (4-4)^2 + (4-7)^2]/3 = 6

Rest will be same as Case 1, 2 and 3

So SD(M) < SD(N) for every case
Sufficient

Answer - B

Posted from my mobile device

Hi avigutman - i said what i said in the yellow highlight because standard deviation is a 'spread' from the mean

Here was my thinking

Set A has more elements that are away from the mean

When it comes to (i), (ii) and (iii) -- there are fewer elements (within the same range) away from the mean. Thus I thought the SD must be smaller too.

I don't see how one can conclude without calculating the variance to know if
(a) is the SD higher or lower between Set A and Set (i) or Set (ii) or Set (iii)
(b) how much does SD move if you eliminate elements (is the movement a big change or a small change) ?

I thought (S2) was not sufficient BECAUSE i thought this maybe was going on

Perhaps in set (i) -- the SD is higher than the SD of Set (A)
whereas in
set (ii) and set (iii) -- the SD is lower than the SD of Set A

Not sure

jabhatta2 Standard deviation is not a 'spread' from the mean. Rather, it's the 'average spread per term', so, conceptually, imagine adding up all the deviations from the mean and dividing that sum by the number of terms (this is not going to give you the true SD, but it's an approximation that works great for GMAT purposes).
How I think through statement (2) in this problem:
we have the same range, but fewer terms. This means that the terms that were removed are terms whose deviations were smaller, while the terms with the greatest deviations (the smallest and largest terms) remain. Therefore, on average, the SD must increase.

Edited to add an analogy:
Class M has seven students, whose heights are all different. Three of those students are chosen to form Group N. Is the average height of the students in Group N greater than the average height of the students in Class M?
(1) ...
(2) The two tallest students in class M are members of Group N.

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