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Is the standard deviation of ages of students in class A

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Is the standard deviation of age of students in class A grea [#permalink] New post 03 Jun 2008, 00:01
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Is the standard deviation of age of students in class A greater than the standard deviation of the age of students in class B?

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Re: DS: Class A & Class B Standard Deviation [#permalink] New post 03 Jun 2008, 00:15
E for me! there is not enough information!

Statement1 and 2 provide no relation btw average age and each, individual age, so you can not figure out exactly SD!
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Re: DS: Class A & Class B Standard Deviation [#permalink] New post 03 Jun 2008, 00:39
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C

SD_A>\sqrt{12} and SD_B<=\sqrt{6}

So, SD_A>\sqrt{12}>\sqrt{6}>=SD_B

SD_A>SD_B
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Re: DS: Class A & Class B Standard Deviation [#permalink] New post 03 Jun 2008, 02:44
Expert's post
It should be 12 and 6 instead of sqr(12) and sqr(6), but it does not influence on reasoning.
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Re: DS: Class A & Class B Standard Deviation [#permalink] New post 03 Jun 2008, 14:30
OA is C

My math concepts around statistics and standard deviation are weak. Can you share some resources for improvement?
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Re: DS: Class A & Class B Standard Deviation [#permalink] New post 03 Jun 2008, 17:26
zeenie wrote:
walker wrote:
C

SD_A>\sqrt{12} and SD_B<=\sqrt{6}

So, SD_A>\sqrt{12}>\sqrt{6}>=SD_B

SD_A>SD_B


Yep. C it is


If More detail is better! :lol: thanks!
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Re: DS: Class A & Class B Standard Deviation [#permalink] New post 03 Jun 2008, 19:29
Expert's post
sondenso wrote:
If More detail is better! :lol: thanks!


Thanks you, you force me to think more carefully :)


fast (guessing) way: the more is difference between values, the more SD. In other words SD corresponds to dispersion of data. Taking both condition, we can see that dispersion of students of A class is obviously less than that of B class. So, C

usual way:

SD=\sqrt{\frac{\sum{(x-x_{av})^2}}{n}}

1) first condition says that for class A |x_j-x_i|>12

Additionally, we can states that minimum SD is (when x_{av} is evenly between x_i and x_j)

SD_{Amin}>\sqrt{\frac{({x_j}-x_{av})^2+({x_i}-x_{av})^2}{2}}=\sqrt{\frac{6^2+6^2}{2}}=6

SD_{Amin}>6

2) second condition says that for class B |x_j-x_i|<=6

Additionally, we can states that maximum SD is (when x_{av} is close to one of x_i or x_j)

SD_{Bmax}<\sqrt{\frac{({x_j}-x_{av})^2+({x_i}-x_{av})^2}{2}}=\sqrt{\frac{6^2+0^2}{2}}=\frac{6}{\sqrt{2}}

SD_{Bmin}<\frac{6}{\sqrt{2}}

1)&2) Combine two conditions:

SD_{A}=>SD_{Amin}>6>\frac{6}{\sqrt{2}}>SD_{Bmin}>=SD_{B}

SD_{A}>SD_{B}
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Re: DS: Class A & Class B Standard Deviation [#permalink] New post 03 Jun 2008, 21:00
walker wrote:
you force me to think more carefully

:P

Walker, one more: How do you come from "statement1: the difference between the ages of any two students in class A is always more than 1 year" to :
walker wrote:
1) first condition says that for class A
, I mean |xj-xi|>12


One more: :lol: I chose E because I follow one rule from Gmatclub that if we dont know exactly specific each age, we can have no conclusion about SD!
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Re: DS: Class A & Class B Standard Deviation [#permalink] New post 03 Jun 2008, 21:12
Expert's post
We have to choose the same units for age to make SD compatible. I choose months and, therefore, |xj-xi|>12

sondenso wrote:
One more: :lol: I chose E because I follow one rule from Gmatclub that if we dont know exactly specific each age, we can have no conclusion about SD!


Think about SD as an average difference between an average value and other data. The problem directly says about differences.
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Re: DS: Class A & Class B Standard Deviation [#permalink] New post 04 Jun 2008, 00:01
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snoor wrote:
My math concepts around statistics and standard deviation are weak. Can you share some resources for improvement?


I think GMAT is far far away from statistics. For many people who did not study statistics SD seems to be a mysterious feature from very difficult mathematical jungle of statistics. But GMAT does not go so deeply. Think about SD as an average deviation of data from an average value and remember the formula.
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Re: DS: Class A & Class B Standard Deviation [#permalink] New post 05 Jun 2008, 08:41
Agree walked I answered this in under 2 mins by simply drawing two bell curves and saw one group was closer to the mean than another. C was suff...
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Is the standard deviation of ages of students in class A [#permalink] New post 25 Mar 2011, 08:48
CLASS AVERAGE AGE NO.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 age of students in class B ?

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

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

Last edited by Bunuel on 10 Jul 2013, 12:46, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: KAPLAN TEST QUES [#permalink] New post 25 Mar 2011, 09:06
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punyadeep wrote:
Q))
CLASS AVERAGE AGE NO.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 age of students in class B ?

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

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

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
EITHER statement BY ITSELF is sufficient to answer the question.
Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.


(1) The difference between the ages of any two students in class A is always more than 1 year.
INSUFFICIENT. No mention about class B here.

(2) No student in class B is more than 6 months older than any other student.
INSUFFICIENT. No mention about class A here.

Combing both and assuming that both classes have at least 2 students;
We can see that the standard deviation will increase with every student added for class A and the the deviation will decrease with every student added to class B.

Maximum standard deviation for class B can be somewhere around 0.25 years or 3 months and the minimum standard deviation of class B will be somewhere around 0.5 years.

Thus Std Dev(A) > Std Dev(B)

Sufficient.

Ans: "C"
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Re: KAPLAN TEST QUES [#permalink] New post 25 Mar 2011, 09:07
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punyadeep wrote:
Q))
CLASS AVERAGE AGE NO.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 age of students in class B ?

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

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

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
EITHER statement BY ITSELF is sufficient to answer the question.
Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.


C.

Statement 1 alone is not enough. It doesn't give any indication what the differences of ages in class B are.
For example, everyone in class B could be exactly the same age, or everyone in class B could be 2 years apart.

Statement 2 alone is not enough. It doesn't give any indication what the difference of ages in class A are.
Everyone in class A could be exactly the same age, or everyone in class A could be 2 years apart.

If you combine them, you know that A has a significantly larger standard deviation in ages.
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Re: KAPLAN TEST QUES [#permalink] New post 26 Mar 2011, 17:09
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A - a1, a2, a3,a4,a5,a6

Total A = 90

a1-a2 > 12 months and so on, but no information about B, so insufficient.

B - b1,b2,b3,b4,b5... b12

Total B = 192

b1-b2 <= 6 month , but no information a bout A, so insufficient


From (1) and (2) together, the difference between elements of set A is > the difference between elements of Set B, and the denominator in A is < the denominator in set B for the Std Dev formula.

=> Std Dev A > Std Dev B

So answer - C.
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Re: KAPLAN TEST QUES [#permalink] New post 26 Mar 2011, 18:21
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standard deviation is the underroot of the (sum of squares of the difference b/w the quantity and the mean divided by the no of quantities).
now since the statement 1 doesnt mentions abt B and stat. 2 doesnt mentions abt A. hence they both are insufficient.

on combining the two statements do we get the full information to make a comparison?
YES.

also keep in mind that the diff b/w the mean and the quantity in B wud always be less than 1 and therefore its square wud be much more smaller. :twisted: :twisted: :twisted:
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Re: KAPLAN TEST QUES [#permalink] New post 04 Nov 2011, 05:52
Thanks for the explanations.
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Re: CLASS AVERAGE AGE NO.OF STUDENTS A 15 YEARS 6 B 16 YEARS 12 [#permalink] New post 04 Jan 2012, 01:28
Question: Is SD(A) > SD(B)?

Statement 1: No mention of students in class B. Thus INSUFFICIENT

Statement 2: No mention of students in class A. Thus INSUFFICIENT

Combining the two statements, we can see that the difference between the ages of two students in class A is considerably larger than the difference between the ages of any two students of class B.
As we know that SD is a measure of the compactness within the elements of a set, we can infer that the elements of Set A are more dispersed than are the elements of Set B. Thus, we know that SD (A) > SD (B). SUFFICIENT.

Answer: C
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Re: CLASS AVERAGE AGE NO.OF STUDENTS A 15 YEARS 6 B 16 YEARS 12 [#permalink] New post 04 Jan 2012, 01:55
C.. Appreciate Fluke suggestion
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Re: DS: Class A & Class B Standard Deviation [#permalink] New post 11 Nov 2013, 12:37
walker wrote:
sondenso wrote:
If More detail is better! :lol: thanks!


Thanks you, you force me to think more carefully :)


fast (guessing) way: the more is difference between values, the more SD. In other words SD corresponds to dispersion of data. Taking both condition, we can see that dispersion of students of A class is obviously less than that of B class. So, C

usual way:

SD=\sqrt{\frac{\sum{(x-x_{av})^2}}{n}}

1) first condition says that for class A |x_j-x_i|>12

Additionally, we can states that minimum SD is (when x_{av} is evenly between x_i and x_j)

SD_{Amin}>\sqrt{\frac{({x_j}-x_{av})^2+({x_i}-x_{av})^2}{2}}=\sqrt{\frac{6^2+6^2}{2}}=6

SD_{Amin}>6

2) second condition says that for class B |x_j-x_i|<=6

Additionally, we can states that maximum SD is (when x_{av} is close to one of x_i or x_j)

SD_{Bmax}<\sqrt{\frac{({x_j}-x_{av})^2+({x_i}-x_{av})^2}{2}}=\sqrt{\frac{6^2+0^2}{2}}=\frac{6}{\sqrt{2}}

SD_{Bmin}<\frac{6}{\sqrt{2}}

1)&2) Combine two conditions:

SD_{A}=>SD_{Amin}>6>\frac{6}{\sqrt{2}}>SD_{Bmin}>=SD_{B}

SD_{A}>SD_{B}


Aren't these SD supposed to be multiplied by the number of persons?
The SD is not just the min and max.... it's from all students, no?
Re: DS: Class A & Class B Standard Deviation   [#permalink] 11 Nov 2013, 12:37
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