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# Set B ={4, 6, 8, 10, 12, 14, 16, 18, 20, 22} Set S is formed

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CEO
Joined: 15 Aug 2003
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Set B ={4, 6, 8, 10, 12, 14, 16, 18, 20, 22} Set S is formed [#permalink]  09 Oct 2003, 14:39
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Set B ={4, 6, 8, 10, 12, 14, 16, 18, 20, 22}

Set S is formed by taking 8 numbers from Set B.

Whats the standard deviation of S?

a) The mean of S is the same as the mean of B

b) S does not include 22
SVP
Joined: 03 Feb 2003
Posts: 1607
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Kudos [?]: 84 [0], given: 0

(1) S can be {6, 8, 10, 12, 14, 16, 18, 20} only; it has a unique SD
(2) S can be {6, 8, 10, 12, 14, 16, 18, 20} or {4, 6, 8, 10, 12, 14, 16, 18} they have the same SD, for they both equally distributed around their means.

Thus D.
Intern
Joined: 17 Sep 2003
Posts: 31
Location: Mumbai,India
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Kudos [?]: 0 [0], given: 0

My working:-

A) The mean of Set B is 13

Therefore, for same mean the sum of set S should be 13x8=104 which means we need to remove two nos with sum 26 from B which has following sets:-
(4,22);(6,20);(8,18);(10,16);(12,14) Therefore a does not give a unique answer

B) S can be formed by removing 22 and any other no. from set B. Therefore, it is also not sufficient to tell SD

Combining A and B we have the set S as Set B- (4,22) . Hence C
_________________

Jazzy bee

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Joined: 15 Aug 2003
Posts: 3467
Followers: 61

Kudos [?]: 718 [0], given: 781

stolyar wrote:
(1) S can be {6, 8, 10, 12, 14, 16, 18, 20} only; it has a unique SD
(2) S can be {6, 8, 10, 12, 14, 16, 18, 20} or {4, 6, 8, 10, 12, 14, 16, 18} they have the same SD, for they both equally distributed around their means.

Thus D.

Stolyar

1 is not sufficient we also have S = { 4,8,10,12,14,16,18,22 }
2...i am not sure if its true.. we also have { 4,8,10,12,14,16,18,20 }

I think Jasdeep makes a lot of sense ..nice way to solve the problem
C should be the correct answer

thanks
praetorian
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Joined: 03 Feb 2003
Posts: 1607
Followers: 7

Kudos [?]: 84 [0], given: 0

Jasdeep wrote:
My working:-

A) The mean of Set B is 13

Therefore, for same mean the sum of set S should be 13x8=104 which means we need to remove two nos with sum 26 from B which has following sets:-
(4,22);(6,20);(8,18);(10,16);(12,14) Therefore a does not give a unique answer

B) S can be formed by removing 22 and any other no. from set B. Therefore, it is also not sufficient to tell SD

Combining A and B we have the set S as Set B- (4,22) . Hence C

How could I be so stupid?
I had a bad day or grow older. Or dealt with the LSAT too long.

(1) agree. removing symmetrical pairs keeps the mean the same
(2) agree. 22 and any other number are out, leaving many different deviations.

combine: S is unique {6, 8, 10, 12, 14, 16, 18, 20} with a unique SD.
agree with C.
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# Set B ={4, 6, 8, 10, 12, 14, 16, 18, 20, 22} Set S is formed

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