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

Question

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

KarishmaB · Accepted Answer

Can you please post your source of this question?
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

BrentGMATPrepNow · Answer

----ASIDE-------------------------

For the purposes of the GMAT, it's sufficient to think of Standard Deviation as the   Average Distance from the Mean  . 
Here's what I mean:

Consider these two sets: Set A {7,9,10,14} and set B {1,8,13,18}
The mean of set A = 10 and the mean of set B = 10
How do the Standard Deviations compare? Well, since the numbers in set B deviate the more from the mean than do the numbers in set A, we can see that the standard deviation of set B must be greater than the standard deviation of set A.

Alternatively, let's examine the Average Distance from the Mean for each set.

Set A {7,9,10,14}  
Mean =  10 
7 is a distance of 3 from the mean of  10 
9 is a distance of 1 from the mean of  10 
10 is a distance of 0 from the mean of  10 
14 is a distance of 4 from the mean of  10 
So, the average distance from the mean = (3+1+0+4)/4 =  2

B {1,8,13,18}  
Mean =  10 
1 is a distance of 9 from the mean of  10 
8 is a distance of 2 from the mean of  10 
13 is a distance of 3 from the mean of  10 
18 is a distance of 8 from the mean of  10 
So, the average distance from the mean = (9+2+3+8)/4 =  5.5

IMPORTANT: I'm  not  saying that the Standard Deviation of set A equals 2, and I'm  not  saying that the Standard Deviation of set B equals 5.5 (They are reasonably close however).

What I am saying is that the average distance from the mean can help us see that the standard deviation of set B must be greater than the standard deviation of set A. 
More importantly, the average distance from the mean is a useful way to think of standard deviation. This model is a convenient way to handle most standard deviation questions on the GMAT.

------NOW ONTO THE QUESTION!!!---------------

So, for this question, we have:

Mean of set A = 70 
Mean of set B = 0 
Mean of set C = 40

100 is furthest away from the mean of 0 in set B, so this will cause the GREATEST change in standard deviation.
100 is next furthest away from the mean of 40 in set C, so this will cause the 2nd greatest change in standard deviation.
100 is closest to the mean of 70 in set A, so this will cause the LEAST change in standard deviation.

Answer: E

Cheers,
Brent

RELATED VIDEO FROM OUR COURSE
 https://www.youtube.com/watch?v=FbAlIjw2fwY

gmat1220 · Answer

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

Initially SD(A) > SD(B) > SD(C). When 100 is added. The range of A is unchanged, so least change. But to calculate any useful relationship between the modified A Vs B we have to know the fact that B contains negative numbers. So we will get new sds as follows SD(B) > SD(C) > SD(A). Pls verify this reasoning.

To prove this inference let me calculate change in mean for sets B and C -
m(B) changes by (100 - 0)/6 = 50/3 = 16.67 hence the new mean of set B is 16.67 + Old mean = 16.67
m(C) changes by (100 - 40)/6 = 10. Hence the new mean of set C is 10 + Old mean = 10 + 40 = 50

Now the new distances from their respective means of set B (mean 16.67) and set C (mean 50)
B = {36.67, 26.67,16.67, 6.67, 3.33, 83.33}
C = {20,15,10,5,0,50}

Hence SD(B) > SD(C) > SD (A). Answer E. So this is not a 120 sec question. How to save time ?

Amateur · Answer

Hey Karishma.... I am stuck.... I dealt with it in another way..... for any series.... If we draw Gaussian curve.... https://upload.wikimedia.org/wikipedia/c ... am.svg.png the min and the max are 8sigma interval difference... where sigma is standard deviation.... in other terms... min and max are 4sigma intervals away from mean.... So if we calculate sigma for the sets as max-min/8.... Set A will be 110-30/8 which is 80/8 Set B will be 100-(-20)/8 which is 120/8 Set C will be 100-30/8 which is 70/8 So in order of highest to lowest wouldn't it be B,A,C? Where am I going wrong? thank you

KarishmaB · Answer

First of all, this is not a normal distribution. In a normal distribution, the values are concentrated around the mean (as is obvious from the normal distribution curve). You cannot calculate the SD of these sets based on the ND curve.
Secondly, you have to order them in terms of the absolute increase  in their standard deviation, not in terms of their new SD.

aalriy · Answer

Hi Karishma,

I can understand that by adding 100 to the three sets the extent to which the S.D changes is based on the absolute difference b/w the mean and 100. But based on this, how can you conclude that the new SD will be in B, C & A order??

Thanks.

KarishmaB · Answer

Notice that the denominator in the calculation of SD will be the same in the case of all the 3 sets (since they all have 5 elements each). When you add 100 to each one of them, they will have 6 elements each and hence the denominator will still stay the same.

In case of set B, the numerator increases by 100^2 (before you take the root)
In case of set C, the numerator increases by 60^2 (before you take the root)
In case of set A, the numerator increases by 30^2 (before you take the root)
So in absolute terms, B will see the most effect and A will see the least. You can look at the actual calculation to understand exactly why this happens. The formula for SD is discussed in the first post below.

Sachin9 · Answer

Hi Karishma,
From what I can infer, it seems that the order of 'increase in the SD' and the order of 'new SD' will always be the same ..
Please correct me if its incorrect.
Regards,
Sach

KarishmaB · Answer

Actually no, that may not be the case.

The increase in SD depends on the distance between the number added (100 here) and the mean. Set B has the smallest mean (0) so it is farthest from 100 hence it will see maximum increase. 
The 'new SD' depends on the difference between all the elements (including the new one) and the mean. If the rest of the numbers are very close to the mean, it is certainly possible that the new SD does not have the same ordering.

e.g.

Set A = {-1, 0, 1}
Set B = (0, 20, 40}
If you add another number say, 30, the increase in SD of set A will be substantial because 30 is far from mean 
but increase in SD of set B will not be very much.
Nevertheless, new SD of set A will be less than the new SD of set B.

Sachin9 · Answer

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..

KarishmaB · Answer

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.

Sachin9 · Answer

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..

KarishmaB · Answer

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.

Daddydekker · Answer

5 sec. approach:

SD is the deviation of the mean to the smallest and biggest number in the set. So if a number is added inside the boundaries of that set, there will be no changes in the SD of that set. So no absolute increase will occur in set A.

A has to be the set with the smallest increase, what's bigger than 0 right? And hence E is the only right answer.

Correct me if I'm wrong

xiaoxueren · Answer

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";

KarishmaB · Answer

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}

kanikab · Answer

VeritasKarishma ,  Bunuel

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

KarishmaB · Answer

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.

ScottTargetTestPrep · Answer

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.

Answer: E

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

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GMAT Problem Solving (PS) Questions

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

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