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Standard Deviation Un-changed

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Standard Deviation Un-changed

by Uva@90 » Fri Nov 08, 2013 8:20 pm
Which of the following has the same standard deviation as {s,r,t}?
I. {r-2, s-2, t-2}
II. {0, s-t, s-r}
III. {|r|, |s|, |t|}
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

OA is D

But I feel answer should be E
Here is My proof:

If we add or subtract a constant to each term in a set Mean will increase or decrease by the same constant SD will not change.
So Set I SD will not change.


Changing the signs of the element of a set has no effect on SD

So set III SD will not change

hence answer should be E

Please explain me which is the right answer.

Thanks in advance

Regards,
Uva.
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by Brent@GMATPrepNow » Fri Nov 08, 2013 10:16 pm
Uva@90 wrote:Which of the following has the same standard deviation as {s,r,t}?
I. {r-2, s-2, t-2}
II. {0, s-t, s-r}
III. {|r|, |s|, |t|}
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
Hi Uva@90,

You've noted the key concept for this question: If we add or subtract a constant to each term in a set, the standard deviation will not change.
For example, set A = {1, 7, 19}
Let's say we add 8 to each value we get set B {9, 15, 27}
The standard deviation of sets A and B are equal.

Now let's examine the 3 sets.

I. {r-2, s-2, t-2}
Here, we are subtracting 2 from each value in the set {s, r, t}, so the standard deviation of this new set must be the same as the standard deviation of the original set.
Aside: or we could say that we're adding -2 to each value in the set
So, THIS SET WORKS

II. {0, s-t, s-r}
Let's start with the set {s, r, t}
If we multiply the values by -1, the new set, {-s, -r, -t}, will have the same standard deviation as the original set {since the values are still spread out the same}.
Now take {-s, -r, -t} and add s to every value to get {-s + s, -t + s, -r + s}, which simplifies to be {0, s-t, s-r}
Since we added s to every value in the set, the new set must have the same standard deviation of the original set.
So, THIS SET WORKS

At this point, we don't need to check the 3rd set. Since I and II both work, the correct answer must be D

Having said that, let's check III for "fun"

III. {|r|, |s|, |t|}
The standard deviation of this set need not be the same as the standard deviation of {r, s, t}

Consider this example: r = 1, s = -1 and t = -1
So, the standard deviation of {|r|, |s|, |t|} equals zero, but the standard deviation of {r, s, t} does NOT equal zero.

So, this set does not work.

Cheers,
Brent
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by Uva@90 » Fri Nov 08, 2013 10:23 pm
Brent@GMATPrepNow wrote:
Uva@90 wrote:Which of the following has the same standard deviation as {s,r,t}?
I. {r-2, s-2, t-2}
II. {0, s-t, s-r}
III. {|r|, |s|, |t|}
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
Hi Uva@90,

You've noted the key concept for this question: If we add or subtract a constant to each term in a set, the standard deviation will not change.
For example, set A = {1, 7, 19}
Let's say we add 8 to each value we get set B {9, 15, 27}
The standard deviation of sets A and B are equal.

Now let's examine the 3 sets.

I. {r-2, s-2, t-2}
Here, we are subtracting 2 from each value in the set {s, r, t}, so the standard deviation of this new set must be the same as the standard deviation of the original set.
Aside: or we could say that we're adding -2 to each value in the set
So, THIS SET WORKS

II. {0, s-t, s-r}
Let's start with the set {s, r, t}
If we multiply the values by -1, the new set, {-s, -r, -t}, will have the same standard deviation as the original set {since the values are still spread out the same}.
Now take {-s, -r, -t} and add s to every value to get {-s + s, -t + s, -r + s}, which simplifies to be {0, s-t, s-r}
Since we added s to every value in the set, the new set must have the same standard deviation of the original set.
So, THIS SET WORKS

At this point, we don't need to check the 3rd set. Since I and II both work, the correct answer must be D

Having said that, let's check III for "fun"

III. {|r|, |s|, |t|}
The standard deviation of this set need not be the same as the standard deviation of {r, s, t}

Consider this example: r = 1, s = -1 and t = -1
So, the standard deviation of {|r|, |s|, |t|} equals zero, but the standard deviation of {r, s, t} does NOT equal zero.

So, this set does not work.

Cheers,
Brent
Brent,
You made it soo clear.
Thanks a ton :)

Regards,
Uva.
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by GMATGuruNY » Sat Nov 09, 2013 4:09 am
Which of the following has the same standard deviation as {s,r,t}?
I. {r-2, s-2, t-2}
II. {0, s-t, s-r}
III. {|r|, |s|, |t|}
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
Standard deviation describes the SPREAD of the data.

Since statement III includes absolute value, test a random combination of positive and negative values.
Let s=-2, r=-1, and t=2.
Here, there is a distance of 1 (between s and r) and a distance of 3 (between -1 and 2).
Any statement that yields a distance of 1 and a distance of 3 will have the same standard deviation.

I: r-2, s-2, t-2
-1-2, -2-2, 2-2
-3, -4, 0
-4, -3, 0.
Here, there is a distance of 1 (between -4 and -3) and a distance of 3 (between -3 and 0).
Eliminate B and C, which do not include statement I.

II: {0, s-t, s-r}
0, -2-2, -2-(-1)
0, -4, -1
-4, -1, 0.
Here, there is a distance of 1 (between -1 and 0) and a distance of 3 (between -4 and -1).
Eliminate A and E, which do not include statement II.

The correct answer is D.

Statement III: {|r|, |s|, |t|}
|-1|, |-2|, |2|
1, 2, 2.
Here, there is a distance of 0 (between 2 and 2) and a distance of 1 (between 1 and 2).
Since statement III does not yield a distance of 1 and a distance of 3, it will not have the same standard deviation as {s, r, t}.
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