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# Sets A, B and C are shown below. If number 100 is included

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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]
xiaoxueren wrote:
vjsharma25 wrote:
Sets A, B and C are shown below. If number 100 is included in each of these sets, which of the following
represents the correct ordering of the sets in terms of the absolute increase in their standard deviation, from
largest to smallest?
A {30, 50, 70, 90, 110}, B {-20, -10, 0, 10, 20}, C {30, 35, 40, 45, 50}
(A) A, C, B
(B) A, B, C
(C) C, A, B
(D) B, A, C
(E) B, C, A

As per me answer should be .
Wants to confirm or reject the OA.

It is a Veritas Prep Book X question for which the OA given is (E). The explanation clearly explains you why the answer is E.
You don't have to calculate anything. SD measures the distance between each element and mean. If a new element is added which is far away from the mean, it will distort the mean more than if it were added close to the mean.
The means of the 3 sets are 70, 0 and 40.
100 is farthest from 0 so it will change the SD of set B the most (in terms of absolute increase). It is closest to 70 so it will change the SD of set A the least. Hence answer is B, C, A

Dear VeritasPrepKarishma,
this is what i found in your post
"If you notice, we have seen two different cases (case 4 and case 5)– in one of them SD increases when you add two numbers to the set and in the other, SD decreases.
Case 4: S = {1, 3, 5} or T = {1, 1, 3, 5, 5} T has higher SD. It has two extra numbers far from the mean
Case 5: S = {1, 3, 5} or T = {1, 3, 3, 5} The standard deviation (SD) of T will be less than the SD of S
So how do you decide whether SD will increase or decrease? Say, what happens in case S = {3, 4, 5, 6, 7} and T = {3, 4, 4, 5, 6, 6, 7}? "
you also said that"If a new element is added which is far away from the mean, it will add much more to the deviations than if it were added close to the mean."

so in case 4: S = {1, 3, 5} ;mean =3 ; the difference of each number of S to mean is 2
T = {1, 1, 3, 5, 5} ; T =3; T added two numbers (1,5) the difference of 1 to the mean(T) of 3 is 2 ; the difference from 5 to the mean(T) of 3 is 2; my question is why It has two extra numbers far from the mean";

When we add numbers far from the mean, they add much more to the numerator and relatively less to the denominator. Though in this question, T does end up having the same SD as S (and that is the caveat I mentioned on my post).

But this is the general concept. So a number added at extreme {1, 3, 5, 100} will increase the SD much more than say {1, 3, 5, 6}
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]

Hi. I solved this Q by a different method, not sure if this is correct or a fluke. Can you please confirm?

Initial SD - A > B > C because of their range (80 >40 >20)
New SD - B > A > C becasue of the new range (120 > 80 > 70).

Since the Q is asking the absolute increase in their standard deviation, I subtracted the new and old SD which led to B (120 - 40)> C(70-20) >A(80-80).

Thanks
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]
kanikab wrote:

Hi. I solved this Q by a different method, not sure if this is correct or a fluke. Can you please confirm?

Initial SD - A > B > C because of their range (80 >40 >20)
New SD - B > A > C becasue of the new range (120 > 80 > 70).

Since the Q is asking the absolute increase in their standard deviation, I subtracted the new and old SD which led to B (120 - 40)> C(70-20) >A(80-80).

Thanks

In this question, range does help us since it is related to the SD in the same way. 100 is the greatest number in each set and the sets are evenly distributed around the mean so 100 increases the range in the same way in which it increases the SD. But for most SD questions, I wouldn't relate range to SD since SD depends on lots of other attributes too.
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]
vjsharma25 wrote:
Sets A, B and C are shown below. If number 100 is included in each of these sets, which of the following represents the correct ordering of the sets in terms of the absolute increase in their standard deviation, from largest to smallest?

A {30, 50, 70, 90, 110}, B {-20, -10, 0, 10, 20}, C {30, 35, 40, 45, 50}

(A) A, C, B
(B) A, B, C
(C) C, A, B
(D) B, A, C
(E) B, C, A

In general, adding a number to a set that is farthest from the mean of the set will increase the standard deviation the most. Therefore, we need to determine the mean of each set.

We can see that each set is evenly spaced, so the mean is also the median. Therefore, the mean of set A is 70, set B is 0, and set C is 40. We see that 100 is furthest from 0 and closest to 70; therefore, the standard deviation will increase the most in set B and the least in set A.

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Sets A, B and C are shown below. If number 100 is included [#permalink]
Sachin9 wrote:
VeritasPrepKarishma wrote:
Sachin9 wrote:
I guess the new SD of A will be more than the new SD of B..
Numbers in A would be more dispersed than those in B..

No. You can use a fin calc to find that the SD of A is 13 and that of B is 14.8. The difference isn't much but still the new SD of A is less than the new SD of B. As I said, what matters is that how far apart are all the elements from the mean in case of new SD. One element can have a huge impact but it still may not be sufficient. So you cannot infer that the new SD will be in the same order.

Thanks a lot, Karishma..

From what I understand,highest effect would depend on how far is the new no. from the mean in all the sets..

and Actual order of SD among sets would depend on how dispersed all elements are from the mean..

Yes, that's correct.
'Change' in SD depends on how far the new no is from the mean. If the new no is close to the mean, the change in SD is very little because it adds very little dispersion to the scenario. If the new no is far from the mean, the change in SD is significant because it adds a lot more dispersion. The mean changes, it becomes farther than the previous mean and hence overall dispersion in a lot higher.

Actual SD depends in big part on how the previous numbers were dispersed around the mean. So it is hard to say what the new order will be based on just the new no. and the previous mean.

Whenever you add a new number to a set -- check how far this new number (in this case 100) is to the pre-existing mean to compare how much the SD will change.

Question : When you add a new number to the set (Set A / Set B or Set C), won't all of them have new averages ?

Thus, shouldn't you be re-calculating all of the absolute distances from this new mean (post addition of 100) to determine the change in SD ? It's quite possible that this new mean may be closer to other elements in the set
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]
jabhatta@umail.iu.edu wrote:
Sachin9 wrote:
VeritasPrepKarishma wrote:
Sachin9 wrote:
I guess the new SD of A will be more than the new SD of B..
Numbers in A would be more dispersed than those in B..

No. You can use a fin calc to find that the SD of A is 13 and that of B is 14.8. The difference isn't much but still the new SD of A is less than the new SD of B. As I said, what matters is that how far apart are all the elements from the mean in case of new SD. One element can have a huge impact but it still may not be sufficient. So you cannot infer that the new SD will be in the same order.

Thanks a lot, Karishma..

From what I understand,highest effect would depend on how far is the new no. from the mean in all the sets..

and Actual order of SD among sets would depend on how dispersed all elements are from the mean..

Yes, that's correct.
'Change' in SD depends on how far the new no is from the mean. If the new no is close to the mean, the change in SD is very little because it adds very little dispersion to the scenario. If the new no is far from the mean, the change in SD is significant because it adds a lot more dispersion. The mean changes, it becomes farther than the previous mean and hence overall dispersion in a lot higher.

Actual SD depends in big part on how the previous numbers were dispersed around the mean. So it is hard to say what the new order will be based on just the new no. and the previous mean.

Whenever you add a new number to a set -- check how far this new number (in this case 100) is to the pre-existing mean to compare how much the SD will change.

Question : When you add a new number to the set (Set A / Set B or Set C), won't all of them have new averages ?

Thus, shouldn't you be re-calculating all of the absolute distances from this new mean (post addition of 100) to determine the change in SD ? It's quite possible that this new mean may be closer to other elements in the set

Yes, the mean will change when we add a number (if we do not add it at the mean) but the impact of one number on the mean will be little (assuming a big set of numbers).
Say we add the number at the mean - then mean doesn't change, there is no extra dispersion of data, but the number of observations increases so SD will actually reduce.
If we add a number far away from the mean, the mean will change slightly, a lot of extra dispersion will be added and the number of observations will increase by 1. Overall impact would be increased mean.

Check out this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2016/0 ... mula-gmat/
It discusses how to differentiate between similar looking situations. GMAT will not give you a situation in which you cannot differentiate appropriately without calculating.
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]
Can someone confirm if by using the range of each set is still relevant than calculating the mean?

A: range = 80
B: range = 0
C: range = 20

So largest impact would be B > C > A given that 100 - range value will be highest for B.
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]
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