I think the answer is D.

I. {r - 2, s - 2, t - 2}
If every element in a set is increased/decreased by a constant, c, the standard deviation does not change. Thus, this set has the same standard deviation as {s, r, t}.

II. {0, s - t, s - r}
We can subtract constant, s, from each of the 3 elements in this set and not change the standard deviation.
The set becomes: {-s, -t, -r}
This is a mirror image of {s,t,r}. The relative positions of {-s,-t,-r} on the number line are the same as the relative positions of {s, r, t}. Thus, standard deviation is the same.

III. {|r|, |s|, |t|}
If {r,s,t} = {-1,1,1}, we have some positive standard deviation.
Then {|r|, |s|, |t|} = {1,1,1}, which has standard deviation = 0.

the answer should be E

I. The spread is consistent with each value so it shouldn't affect the SD
II. We don't know the spread of each value. It could be big or small
III. The absolute value sign measures the distance from 0. So it should be the same.

For Same STD Deviation, both the sets need to have values dispersed in a similar way...
eg Option 1) all values are same as Original delta of 2
Option 2) first value is 0, second is s-t, and then s-r all 3 are differently spread.
Option 3) absolute values of s,r,t are going to have same spread as original set.

So Answer E with option 1 and option 2.


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Absolute can change the value of average from the original so E cannot be the option

IMO A

Dear Bunuel, what’s the answer? Thank you.


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The OA will be automatically revealed on Saturday 14th of April 2018 01:47:19 PM Pacific Time Zone
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The answer should be (D) i.e. Both I AND II

I.
The SD for a set of number remains same if added or subtracted by a constant. So, (r-s, s-2, and t-2 ) has same SD as (r, s, and t).
This is because the spread between the nos. is not changing. Only the set of nos. is shifting 2 units away.

II
(-s, -r, and -t) are just the mirror image of (s, r and t) on the number line. Thus the set (-s, -r and -t) has the same spread as the set (s, r and t).
Thus the SD for (-s, -r and -t) is same as that of (s, r and t).
Also if a set is added by a constant number its SD remains the same.
So, (-s+s, -r+s and -t+s) has same SD that of (-s, -r and -t)
Thus, (0, s-t, s-r) has same SD as that of (s, r, t)

Pick easy consecutive numbers and test. Say {s,r,t} is {-2,0,2} so deviation from mean of 0 is 2

I. {-4,-2,0} deviation from mean of -2 is also 2
II. {0, -2-2, -2-0} or {-4,-2,0}, the same as above
III. {|-2|, |0|, |2|} or {0, 2, 2}, both the mean and deviation have changed

Check this post:
https://www.veritasprep.com/blog/2015/0 ... -the-gmat/

If you add/subtract the same number from each element of a set, the SD does not change.

I. {r - 2, s - 2, t - 2}
Subtracting 2 from each elements will not change the SD.

II. {0, s - t, s - r}

{s - s, s - t, s - r}

Note that whatever the SD of {s, r, t}, the same will be the SD of {-s, -r, -t} because relative distance between them on the number line does not change (all the numbers are just flipped across the Y axis - positive becomes negative, negative becomes positive).
So, when we add s to each term, we still get the same SD.

III. {|r|, |s|, |t|}

This could change the relative distance between them on the number line if r, s and t all do not have the same sign.
e.g.

______________ (-5)_______________________0______________(3)____(4)

becomes

__________________________________________0______________(3)____(4)____ (5)
Much smaller SD now.

Answer (D)
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