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

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Set B ={4, 6, 8, 10, 12, 14, 16, 18, 20, 22} Set S is formed [#permalink] New post 09 Oct 2003, 15:39
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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
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 [#permalink] New post 10 Oct 2003, 00:19
(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.
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 [#permalink] New post 10 Oct 2003, 06:24
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
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 [#permalink] New post 12 Oct 2003, 23:29
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

stolyar, comments?

thanks
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 [#permalink] New post 12 Oct 2003, 23:44
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? :oops: :toilet :wall
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.
  [#permalink] 12 Oct 2003, 23:44
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Set B ={4, 6, 8, 10, 12, 14, 16, 18, 20, 22} Set S is formed

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