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Set A consists of all prime numbers between 10 and 25; Set B [#permalink]

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14 Mar 2011, 09:05

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Set A consists of all prime numbers between 10 and 25; Set B consists of consecutive even integers, and set C consists of consecutive multiples of 7. If all the three sets have an equal number of terms, which of the following represents the ranking of these sets in an ascending order of the standard deviation?

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

Set A consists of all prime numbers between 10 and 25; Set B consists of consecutive even integers, and set C consists of consecutive multiples of 7. If all the three sets have an equal number of terms, which of the following represents the ranking of these sets in an ascending order of the standard deviation?

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

Set A - {11, 13, 17, 19, 23}; Set B - {5 consecutive even integers}, for example - {12, 14, 16, 18, 20}; Set C - {5 consecutive multiples of 7}, for example - {7, 14, 21, 28, 35}.

Now, the standard deviation of a set shows how much variation there is from the mean, how widespread a given set is. So, a low standard deviation indicates that the data points tend to be very close to the mean, whereas high standard deviation indicates that the data are spread out over a large range of values.

You can see that set C is most widespread and set B is least widespread, so the correct answer is: B, A, C.

the question is from 700-800 level questions.........however there is no source,,,if u wish i can upload the document cos i find this document very good...!!!

SD will be high if the sample is not compact since SD is a measure of compactness.

Mean will fall in the middle of the highest and lowest. So more the distance from the mean - the higher the SD. Assuming all the data is positive the SD will be proportional to the range (highest - smallest)

Set A {11 13 17 19 23} range = 12 Se B {12 14 16 18 20} range = 8 Set C {14 21 28 35 42 } range = 28

the question is from 700-800 level questions.........however there is no source,,,if u wish i can upload the document cos i find this document very good...!!!

Just my 2 cents: there are a lot of "good" documents around.. but least we can do is be careful on our part. If others didnt testify the right answer, I would have missed my dinner!

Edit: I just found out, this question is from Veritas Prep "Statistics and Problem Solving" book and the OA is indeed E.
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the question is from 700-800 level questions.........however there is no source,,,if u wish i can upload the document cos i find this document very good...!!!

@ fluke...two documents iam uploading......they were sent to me by a frnd who scored 750 and he just went through this document for quant twice.......i found it focussed and helpful too...!!

Re: Set A consists of all prime numbers between 10 and 25; Set B [#permalink]

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12 Jul 2013, 10:28

punyadeep wrote:

Set A consists of all prime numbers between 10 and 25; Set B consists of consecutive even integers, and set C consists of consecutive multiples of 7. If all the three sets have an equal number of terms, which of the following represents the ranking of these sets in an ascending order of the standard deviation?

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

Set A \(\to {11,13,17,19,23}\)

Fact: Addition/subtraction of any number from a given set doesn't change the S.D of the given set.

From Set A, subtract 11 across to represent \(Set A^'\) as \(\to {0,2,6,8,12}\). \(S.D of Set A = S.D of Set A^'\)

If you notice closely, Set \(A^'\) is very close to Set B, just that Set B looks like \(\to{0,2,4,6,8}\),i.e. a data-point of 12 is present instead of 4, which makes \(Set A^'\) more spread-out then Set B.

Thus, \(S.D of Set B< S.D of Set A^'\)

Assume Set C\(\to {0,7,14,21,28}\)

S.D represents how far are the data points are from the mean of the set, we can see that Set C has the most spread-out data points and the correct order = B,A,C.

Re: Set A consists of all prime numbers between 10 and 25; Set B [#permalink]

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12 Jul 2013, 10:47

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punyadeep wrote:

Set A consists of all prime numbers between 10 and 25; Set B consists of consecutive even integers, and set C consists of consecutive multiples of 7. If all the three sets have an equal number of terms, which of the following represents the ranking of these sets in an ascending order of the standard deviation?

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

Hi,

few things to keep in mind. =>std deviation depends on distance of each term from mean and number of terms.==>since number of terms are same..we are concerned only about distance of EACH TERM from MEAN.==more the distance more the deviation. =>std deviation is always positive==>so not concerned whether numbers are positive or negative.

SET A==>(11,13,17,19,23)==>avg around 16==>you can see distance of 11,13,17,19,23 is 5,3,1,3,7....1) SET B==>(CONSECUTIVE EVEN INTEGERS...)= ( N-4,N-2,N,N+2,N+4)===>mean=N==>DISTANCE FROM MEAN= 4,2,0,2,4 SET C==>MULTIPLES OF 7 ====N-14,N-7,N,N+7,N+14===>MEAN=N====>DISTANCE FROM MEAN=14,7,0,7,14.

if we compare all distance clearly we can say B<A<C===>HENCE OPTION E
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Re: Set A consists of all prime numbers between 10 and 25; Set B [#permalink]

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14 Oct 2014, 06:24

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Re: Set A consists of all prime numbers between 10 and 25; Set B [#permalink]

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30 Dec 2015, 00:24

gmat1220 wrote:

SD will be high if the sample is not compact since SD is a measure of compactness.

Mean will fall in the middle of the highest and lowest. So more the distance from the mean - the higher the SD. Assuming all the data is positive the SD will be proportional to the range (highest - smallest)

Set A {11 13 17 19 23} range = 12 Se B {12 14 16 18 20} range = 8 Set C {14 21 28 35 42 } range = 28

SD(B) < SD(A) < SD(C)

Please correct me if this reasoning is wrong.

Buneal, please could you provide more info on how the standard deviation is proportional to the range. Is this strategy useful when you are in stressed conditions?

Re: Set A consists of all prime numbers between 10 and 25; Set B [#permalink]

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31 Dec 2016, 10:22

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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