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Director  Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
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For a certain exam, was the standard deviation of the scores  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 63% (01:09) correct 38% (01:11) wrong based on 412 sessions

HideShow timer Statistics For a certain exam, was the standard deviation of the scores for students U, V, W, X, Y, Z less than the standard deviation of the scores for students A B and C?

1. The standard deviation of the scores of students U V and W was less than the standard deviation of the scores of the students A B and C on the exam

2. The standard deviation of the scores of students X Y and Z was less than the standard deviation of the scores of the students A B and C on the exam
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Re: SD of the combined set  [#permalink]

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Good thinking. This is how I will go about analyzing these statements.

1. SD is a measure of compactness, not the measure of the absolute value of the number. Hence if we have two sets of numbers and we want to know the SD of the combined set, then we must know the mean and distance of each data separately.

Suppose {U V W } and { X Y Z } are combined to form a new set { U V W X Y Z } then the SD of the combined set cannot be predicted, even if we know that
SD (U V W) < SD ( A B C)
and
SD (X Y Z) < SD (A B C)

SD(u,v,w,x,y,z) < SD(a,b,c) ---- this cannot be inferred.

When I see SD(U, V, W) < SD(A, B, C),
I think of this case:
U=V=W=10 so SD(X, Y, Z) = 0
SD(A, B, C) has some positive value. Say A = 1, B = 2, C = 3

When I see SD(X, Y, Z) < SD(A, B, C),
I think of this case:
X=Y=Z=20 so SD(X, Y, Z) = 0
SD(A, B, C) has some positive value. Say A = 1, B = 2, C = 3

Now, SD(U, V, W, X, Y, Z) = SD(10,10,10, 20, 20, 20) which is more than SD (A, B, C)
On the other hand, if U = X, SD(U, V, W, X, Y, Z) = SD(10,10,10,10,10,10) = 0 which is less than SD (A, B, C)

So I cannot say anything about the combined SD.

2. On the other hand two sets - { L M N } and { X Y Z } and it is given that -

SD { L M N} > SD ( X Y Z)

If we add numbers to the first set - { P Q R } and then the set becomes { L M N P Q R}. We can infer that -

SD (L M N P Q R) > SD (X Y Z)[/quote]

Let me write down some relations:
SD(10, 11, 12) > SD(10, 11, 11) (less dispersed, closer together)
SD(10, 11, 12) < SD(9, 10, 11, 12, 13) (More dispersed)
SD(10, 11, 12) > SD(10, 11, 11, 11, 12) (Closer together - If this is not apparent, think of the formula of SD. It is root of the sum of the squares of the distance from the mean divided by the number of elements. If you keep adding more numbers at the mean, number of elements keeps increasing but the sum of distance doesn't so SD keeps going down.

So when you add new elements, SD could increase/decrease. Hence, you cannot infer anything in this case either.

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Re: SD of the combined set  [#permalink]

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gmat1220 wrote:
Karishma / fluke / Ian
Can you please verify this reasoning? I hope there are 2 takeaways from this question -

1. SD is a measure of compactness, not the measure of the absolute value of the number. Hence if we have two sets of numbers and we want to know the SD of the combined set, then we must know the mean and distance of each data separately.

Suppose {U V W } and { X Y Z } are combined to form a new set { U V W X Y Z } then the SD of the combined set cannot be predicted, even if we know that

SD (U V W) < SD ( A B C)
and
SD (X Y Z) < SD (A B C)

SD(u,v,w,x,y,z) < SD(a,b,c) ---- this cannot be inferred.

This one is pretty straight forward.
{A,B,C}={1,3,5}
{U,V,W}={1,2,3}
{X,Y,Z}={1000000000,1000000001,1000000002}
OR
{X,Y,Z}={1,2,3}

2. On the other hand two sets - { L M N } and { X Y Z } and it is given that -

SD { L M N} > SD ( X Y Z)

If we add numbers to the first set - { P Q R } and then the set becomes { L M N P Q R}. We can infer that -

SD (L M N P Q R) > SD (X Y Z)

This one is bit trickier.
This is my rough analysis.
When we have 3 terms with a particular range and standard deviation. If we add more numbers within the maximum and minimum value, the new standard deviation will be less.

{1,2,3}: min=1, max=3;
if we add any number within 1 and 3, the std dev will decrease. say, we add "2", std dev will decrease
if we add any number beyond 1 and 3, the std dev will increase. say, we add "0.9" or "3.1", std dev will increase.
Thus, we can't infer this either.

{L M N} > {X Y Z}
L,M,N=0.9,3,5.1
X,Y,Z=1,3,5
If
P,Q,R=3,3,3(Very close to the mean and within 0.9 and 5.1)
{L,M,N,P,Q,R}<{X,Y,Z}

If
P,Q,R=5.1,0.9,0.9(Pretty skewed from the mean and is more than 5.1 and <=0.9)
{L,M,N,P,Q,R}>{X,Y,Z}

I agree this is convoluted!!
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Re: SD of the combined set  [#permalink]

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Karishma / fluke / Ian
Can you please verify this reasoning? I hope there are 2 takeaways from this question -

1. SD is a measure of compactness, not the measure of the absolute value of the number. Hence if we have two sets of numbers and we want to know the SD of the combined set, then we must know the mean and distance of each data separately.

Suppose {U V W } and { X Y Z } are combined to form a new set { U V W X Y Z } then the SD of the combined set cannot be predicted, even if we know that

SD (U V W) < SD ( A B C)
and
SD (X Y Z) < SD (A B C)

SD(u,v,w,x,y,z) < SD(a,b,c) ---- this cannot be inferred.

2. On the other hand two sets - { L M N } and { X Y Z } and it is given that -

SD { L M N} > SD ( X Y Z)

If we add numbers to the first set - { P Q R } and then the set becomes { L M N P Q R}. We can infer that -

SD (L M N P Q R) > SD (X Y Z)
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Re: SD of the combined set  [#permalink]

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let the scores be out of 200.

a,b,c = 40,50,60 SD = approx 10 considering the difference only.

u,v,w = 20,25,30 and x,y,z = 35,40,45 | 105,110,115

now either a or b alone is not sufficient.

a+b sets will be (20,25,30,35,40,45) SD approx 5

(20,25,30,105,110,115) SD >10

Hence E.
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Re: SD of the combined set  [#permalink]

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gmat1220 wrote:
Karishma / fluke / Ian
Can you please verify this reasoning? I hope there are 2 takeaways from this question -

2. On the other hand two sets - { L M N } and { X Y Z } and it is given that -

SD { L M N} > SD ( X Y Z)

If we add numbers to the first set - { P Q R } and then the set becomes { L M N P Q R}. We can infer that -

SD (L M N P Q R) > SD (X Y Z)

Set (p,q,r) can have values such that they are closer to the mean of the set L,M,N.
(L,M,N) = (2,4,6) (P,Q,R) = (4,4,4) The SD decreases.
and similarly (P,Q,R) can have values such that SD will increase.

Here degree of relativeness has to be tested.
Meaning, If SD(L,M,N) > SD (X,Y,Z)

then SD(L,M,N,P,Q,R) may be greater than or lesser than SD (X,Y,Z) depending upon by how much SD(P,Q,R) effects.
Director  Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
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Re: SD of the combined set  [#permalink]

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Ohh so we have to know the data exactly to determine the SD. There are no short cuts I believe. fluke and amit2k9 +1
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Re: SD of the combined set  [#permalink]

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gmat1220 wrote:
Ohh so we have to know the data exactly to determine the SD. There are no short cuts I believe. fluke and amit2k9 +1

thanks for bringing up the discussion gmat1220. It was useful.
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Re: SD of the combined set  [#permalink]

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I'd just add that you really don't need to know much about how the standard deviation will be affected when you enlarge a set. If you take the set {1, 3, 5}, and you add the number 4.5 to the set, it's mathematically quite complicated to see what effect that will have on the standard deviation, mainly because your mean changes, so *all* of the distances change. While you might need to be concerned with this kind of situation in an advanced statistics course, you'll certainly never need to worry about it on the GMAT.

The lone situation you'd want to understand is the one Karishma discusses above: if you enlarge a set by adding new elements which are exactly equal to the mean, that will bring the standard deviation closer to 0, because you are adding elements with a distance of 0 to the mean. So if you have the set {1, 3, 5}, and you add new elements to this set which are equal to 3, that will lower the standard deviation.

As for the question in the original post, the answer is E. One can take two extreme examples to see this easily. Suppose our test is graded out of 100, and that test-takers A, B and C had scores of 49, 50 and 51. Their standard deviation is then some small positive number. We could have either of the following scenarios, using both statements:

* U, V, W, X, Y and Z all received scores of 100 on the test. Their standard deviation is thus 0 when we combine their scores into one set.

* U, V and W received scores of 100 on the test, and X, Y and Z received scores of 0 on the test. Then when we combine their scores into one set, we have the set {0, 0, 0, 100, 100, 100} which has a comparatively large standard deviation, certainly much larger than that of {49,50,51}.
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Re: SD of the combined set  [#permalink]

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My quick and dirty thought process:
When I think about standard deviation I think about variance. How different the scores are from one another.

There's no way I can know for sure whether answers A or B are true by themselves, I think that's fairly obvious. With both assumptions combined, I just created a simple hypothetical situation and realized I still didn't have enough information:

(U,V,W) = (1,2,3)
(X,Y,Z) = (100,101,102)

(A,B,C) = (7,9,11)

From this set up it becomes clear that unless there is a restriction as to what the scores can be, determining the difference in variance between the two groups is impossible.
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Re: For a certain exam, was the standard deviation of the scores  [#permalink]

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Wonderful question gmat1220. Thanks.

Thanks for the explanations Ian, Karishma and fluke. Learnt a very vital lesson here. _________________
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Re: For a certain exam, was the standard deviation of the scores  [#permalink]

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