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Class..............Average Age........# of Students

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Class..............Average Age........# of Students [#permalink] New post 25 Feb 2004, 08:52
Class..............Average Age........# of Students
A..................15 years.....................6
B..................16 years.....................12

Is the standard deviation of ages of students in class A greater than the standard deviation of the ages of students in class B?

(1) The maximum difference between the ages of any two students in class A is 1 year.

(2) No student in class B is more than 6 months older than any other student.

Just do not say C and give the dispersion justificatification. Please, show mathematically if C indeed is the correct answer. Thanks
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 [#permalink] New post 25 Feb 2004, 12:54
I do not know how to give better explanation except that st1) provides information only about class A, and st2) only about class B so in order to compare both classes we need both statements.The rest is the dispersion stuff.
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 [#permalink] New post 25 Feb 2004, 13:10
OE can be found in Kaplan GMAT 2004 book..practice test.

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I am not convinced because [#permalink] New post 25 Feb 2004, 13:12
I am not convinced because the mean and range are given. Here is some related info:

The range is the simplest measure of spread or dispersion: It is equal to the difference between the largest and the smallest values. The range can be a useful measure of spread because it is so easily understood. However, it is very sensitive to extreme scores since it is based on only two values.

The range should almost never be used as the only measure of spread, but can be informative if used as a supplement to other measures of spread such as the standard deviation or semi-interquartile range.
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 [#permalink] New post 25 Feb 2004, 15:10
Class..............Average Age........# of Students
A..................15 years.....................12
B..................16 years.....................12

Is the standard deviation of ages of students in class A greater than the standard deviation of the ages of students in class B?

(1) The maximum difference between the ages of any two students in class A is 1 year.

(2) No student in class B is more than 6 months older than any other student.


I agree with Mantha.

I've incresed the number of students in class A from 6 to 12.

Try Anwering the restated problem and you get the clue!

Last edited by kpadma on 25 Feb 2004, 18:04, edited 2 times in total.
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 [#permalink] New post 25 Feb 2004, 17:16
Kpadma and Mantha, I don't agree with your answers and postings.
Mantha, the range, yes, is informative, but it tells you the spread between extreme values. The standard deviation is a measure of dispersion about the mean. This is very different and will have different implications.

For instance, you can have a series of data with 2 extreme outliers which will affect your range but if all the other values (say 100) are around the mean, then the standard deviation might not change much (ie 1*100, 1*1 and 100*50). On the other hand, you may have a same set of numbers but with a smaller range and whereby all the values gravitate around those range values. In this case, the standard deviation will be larger but the range can be smaller than in the first example (ie 51*75, 51*25). This is the difference between the range and the standard deviation.

Now, what does the question ask you? It basically gives you the standard deviation but gives you NO CLUE about the range. For all we know, in class A, there could be extreme values between 1 and 30 years which will make the average become 15. However, the first premise says that "the maximum difference between the ages of any two students in class A is 1 year" This clearly gives you the standard deviation because if you take ANY TWO student, you will not get anything more than 1 year. Therefore, the range could be between 14.3-15.3 years or could also be between 14.8-15.8 years. How could that be? If the range is between 14.3 and 15.3 it means that the mean is biased upward and there are more students clustered above the 15 years old age. If the range is between 14.8 and 15.8 years, it means that the mean is biased downward and there are more students clustered below the 15 years old age. However, in both cases, the standard deviation, the difference between ANY TWO studends picked, is still 1 year.

Therefore, all we know from the first premise is the standard deviation, not the range. Also, as I said in the original posted question, the number of students DOES NOT affect the standard deviation because when it is given, it will not change whether you have 100 students or 2 students. The first premise gives you the standard deviation of class A and the second premise gives you the standard deviation of class B and you need both to say which class' standard deviation is larger.

In conclusion, in answer to Kpadma's question, even with the increased number of students, the answer will be the same. You still need both statements to answer the problem because we need to know both st. dev. and once again, the range has nothing to do with this question
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I tend to disagree.. [#permalink] New post 25 Feb 2004, 17:48
What is given in the question is Range not SD..

SD cannot be known until you know all the values and the mean..

Can someone post the official explanation verbatim from Kaplan please

Last edited by mantha on 25 Feb 2004, 18:19, edited 1 time in total.
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Re: I tend to disagree.. [#permalink] New post 25 Feb 2004, 17:59
mantha wrote:
What is given in the question is Range not SD..

SD cannot be known until you know all the values and the mean..


Hmmm, right, it is the range but should the SD not also be 1 for the range to be within 1 unit above/below the mean for the average to be 15? For instance, you cannot have 10*16 and 10*18 because the average would be 17. Therefore, the SD has to be about 1 about the average of 15
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 [#permalink] New post 25 Feb 2004, 18:02
Paul: Thank you for such a lively post and debate. Now, I need to at least come up with an argument that has 10 lines. For me, typing is hard. Please excuse my short postings.

My Answer to the modified Question,

Class..............Average Age........# of Students
A..................15 years.....................12
B..................16 years.....................12

Is the standard deviation of ages of students in class A
greater than the standard deviation of the ages of students
in class B?

(1) The maximum difference between the ages of any two students in class A is 1 year.

(2) No student in class B is more than 6 months older than any other student.


I would go with E. (Both statements are not sufficient)


Data given:

Class.......Average age..........#of students.........Range
A ...............15 ...........................12...................1

Case 1: Student 1 = 14.5, student 2 = 15.5 , all others are 15;
SD = 0.5/SQRT(6)

Case 2: Six students are 14.5 and other six are 15.5;
SD = 0.5

The standard deviation is between 0.2 to 0.5 (At least for this data set)

Class......Average age............#of students.........Range
B.................16.............................12.................0.5

Case 1: Student 1 = 15.75, student 2 = 16.25, all others are 16;
SD = 0.25/SQRT(6)

Case 2: Six students are 15.75 and other six are 16.25;
SD = 0.25

The standard deviation is between 0.1 to 0.25.

From the two ranges of standard deviations, we can see
that they do overlap. That proves that RANGE is not sufficient
to find standard deviation.
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 [#permalink] New post 25 Feb 2004, 18:16
kpadma wrote:
Paul: Thank you for such a lively post and debate. Now, I need to at least come up with an argument that has 10 lines. For me, typing is hard. Please excuse my short postings.

My Answer to the modified Question,

Class..............Average Age........# of Students
A..................15 years.....................12
B..................16 years.....................12

Is the standard deviation of ages of students in class A
greater than the standard deviation of the ages of students
in class B?

(1) The maximum difference between the ages of any two students in class A is 1 year.

(2) No student in class B is more than 6 months older than any other student.


I would go with E. (Both statements are not sufficient)


Data given:

Class.......Average age..........#of students.........Range
A ...............15 ...........................12...................1

Case 1: Student 1 = 14.5, student 2 = 15.5 , all others are 15;
SD = 0.5/SQRT(6)

Case 2: Six students are 14.5 and other six are 15.5;
SD = 0.5

The standard deviation is between 0.2 to 0.5 (At least for this data set)

Class......Average age............#of students.........Range
B.................16.............................12.................0.5

Case 1: Student 1 = 15.75, student 2 = 16.25, all others are 16;
SD = 0.25/SQRT(6)

Case 2: Six students are 15.75 and other six are 16.25;
SD = 0.25

The standard deviation is between 0.1 to 0.25.

From the two ranges of standard deviations, we can see
that they do overlap. That proves that RANGE is not sufficient
to find standard deviation.


Oh, your example perfectly makes sense. Well, you now clearly proved your point and it was nice. I misunderstood the premises for being the actual SD as mantha pointed it out.
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 [#permalink] New post 26 Feb 2004, 12:04
Exciting conversation!! Thanks Paul for posting that big post.

Just my 0.02 $,

I think GMAT needs us to remember the concept of dispersion only as far as standard deviation is concerned.

I came up with answer E (not by kpadma's method though).

The answer would be E for original problem as well as for kpadma's modified problem.

We need to read both the statements more closely.

(1) The maximum difference between the ages of any two students in class A is 1 year.

This does not mean that the DISPERSION is actually 1. It could be anywhere between 0 and 12 months. So NOT sufficient

(2) No student in class B is more than 6 months older than any other student.

In this case also, it is not necessary that the dispersion is 6 months. It could be anywhere between 0 to 6 months. NOT SUFFICIENT

BOTH statements TOGETHER: We can not say which class has got greater dispersion.

So the answer is E
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 [#permalink] New post 26 Feb 2004, 12:11
I believe Kaplan answer was C on this controversial question. Someone should email them to correct their mistake :writer
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 [#permalink] New post 26 Feb 2004, 15:29
gmatblast wrote:
I came up with answer E (not by kpadma's method though).

The answer would be E for original problem as well as for kpadma's modified problem.

So the answer is E



Dear GmatBlast,

The more we discuss, the more we learn. The golden rule to
succed in life is that "Never afraid of failures". No one in the
world ever learn to walk before falling a few times. That being
said, don't afraid to post your method even if you think it may be
incorrect. You can always remove the posting, after few days
if it is incorrect.


Could you let us know the way you solved this question?


I think the answer to the original question is C.

For the set A, SD is between 0.288 and 0.5
For the set B, SD is between 0.102 and 0.25.

There is no overlap between these ranges. For any combinations of
ages, SD(A) > SD(B).

Thus, answer is C.

That is why, I changed the problem to prove the misconception.
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 [#permalink] New post 26 Feb 2004, 15:34
kpadma wrote:
gmatblast wrote:
I came up with answer E (not by kpadma's method though).

The answer would be E for original problem as well as for kpadma's modified problem.

So the answer is E



Dear GmatBlast,

The more we discuss, the more we learn. The golden rule to
succed in life is that "Never afraid of failures". No one in the
world ever learn to walk before falling a few times. That being
said, don't afraid to post your method even if you think it may be
incorrect. You can always remove the posting, after few days
if it is incorrect.


Could you let us know the way you solved this question?


I think the answer to the original question is C.

For the set A, SD is between 0.288 and 0.5
For the set B, SD is between 0.102 and 0.25.

There is no overlap between these ranges. For any combinations of
ages, SD(A) > SD(B).

Thus, answer is C.

That is why, I changed the problem to prove the misconception.


Kpadma,

Am I missing something in your message? I have posted my answer with explanation only.

You included only 3 lines of response from my post within the quot. It seems you were not able to see my full post. Can you try again?
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Re: Old one re-posted.. [#permalink] New post 27 Feb 2004, 18:29
mantha wrote:
Class..............Average Age........# of Students
A..................15 years.....................6
B..................16 years.....................12

Is the standard deviation of ages of students in class A greater than the standard deviation of the ages of students in class B?

(1) The maximum difference between the ages of any two students in class A is 1 year.

(2) No student in class B is more than 6 months older than any other student.

Just do not say C and give the dispersion justificatification. Please, show mathematically if C indeed is the correct answer. Thanks


The answer for this problem is C.


1) tells you that the maximum difference is 1 year. I know that some of you disagree, but this means that there are at least two students 1 year apart. (it is like saying the maximum of this set {1,2,8} is 8, you cannot say that maximum of the set is 100). Hence, the set with the minimum SD is 14.5, 15.5, 15, 15, 15, 15 with SD 2.88.

2) tells you that no student is more that 6 months older than another. Hence, the maximum SD would be 15.75 x 6 and 16.25 x 6 with SD of .25. Of course, the minimum SD would be 0 if all of them were 16 ("no student is more than six months older" doesn't means "at least one student must be six month older than another").

Hence, with 1 and 2 known, the SDs do not overlap and the queston can be answered. Thus, the answer is C.
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what is this.. [#permalink] New post 27 Feb 2004, 19:34
I do not understand this:

2) .....
Hence, the maximum SD would be 15.75 x 6 and 16.25 x 6 with SD of .25.
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Re: what is this.. [#permalink] New post 27 Feb 2004, 20:05
mantha wrote:
I do not understand this:

2) .....
Hence, the maximum SD would be 15.75 x 6 and 16.25 x 6 with SD of .25.


The maximum SD would occur if 6 people were 15 years and 9 months old and 6 people were 16 years and 3 months old. The minimum would occur if everybody was exactly 16 years old.
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MBA, Anderson School of Management, UCLA, Class of 1993

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Can this be thought thru in less than 2 min? [#permalink] New post 27 Feb 2004, 20:23
I am just wondering what kind of SD questions the GMAT taker should be familiar with. Can you point or list a few of the SD questions resembling the ones on actual GMAT?

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Re: Old one re-posted.. [#permalink] New post 18 Aug 2004, 05:20
AkamaiBrah wrote:
mantha wrote:
Class..............Average Age........# of Students
A..................15 years.....................6
B..................16 years.....................12

Is the standard deviation of ages of students in class A greater than the standard deviation of the ages of students in class B?

(1) The maximum difference between the ages of any two students in class A is 1 year.

(2) No student in class B is more than 6 months older than any other student.

Just do not say C and give the dispersion justificatification. Please, show mathematically if C indeed is the correct answer. Thanks


The answer for this problem is C.


1) tells you that the maximum difference is 1 year. I know that some of you disagree, but this means that there are at least two students 1 year apart. (it is like saying the maximum of this set {1,2,8} is 8, you cannot say that maximum of the set is 100). Hence, the set with the minimum SD is 14.5, 15.5, 15, 15, 15, 15 with SD 2.88.

2) tells you that no student is more that 6 months older than another. Hence, the maximum SD would be 15.75 x 6 and 16.25 x 6 with SD of .25. Of course, the minimum SD would be 0 if all of them were 16 ("no student is more than six months older" doesn't means "at least one student must be six month older than another").

Hence, with 1 and 2 known, the SDs do not overlap and the queston can be answered. Thus, the answer is C.



AkamaiBrah/KPadma:


When you calculate your standard deviation (e.g. to prove your cases) how do you do it? Do you employ the formulae:

std.deviation = sqrt([sum(x-x*)]/[n-1])?

In your examples how do you come up with stddeviation to prove your case? Did you truly calculate the aforementioned? Is there an "easy way" to approximate standard deviation?
Re: Old one re-posted..   [#permalink] 18 Aug 2004, 05:20
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