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seekmba
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Set A and B are 2 sets of numbers. A has a standard 23 Sep 2010, 18:39
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

56% (01:15) correct 44% (01:49) wrong based on 59 sessions

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Set A and B are 2 sets of numbers. A has a standard deviation 3 and a mean 5. B has the same mean but standard deviation 4. Can we find the standard deviation of the set A U B ? If no, then is it possible with the following
additional information?

(1) Both sets have the same number of members each.
(2) Both sets have distinct members (no number is common to both sets).

I found this on the forum here but did not understand why the OA is A. May be Bunuel can explain this.
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A
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adishail
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Set A and B are 2 sets of numbers. A has a standard 23 Sep 2010, 20:03
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seekmba
Set A and B are 2 sets of numbers. A has a standard deviation 3 and a mean 5. B has the same mean but standard deviation 4. Can we find the standard deviation of the set A U B ? If no, then is it possible with the following
additional information?

(1) Both sets have the same number of members each.
(2) Both sets have distinct members (no number is common to both sets).

I found this on the forum here but did not understand why the OA is A. May be Bunuel can explain this.
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Standard Dev for A = SDa

Standard Dev for B = SDb

Standard Dev for AUB = SDaub

Mean = M

Terms of A = X1, X2, ... Xna (Total terms Na)

Terms of B = Xna+1, Xna+2 , ... Xnb (Total terms Nb)

SDa = \(\sqrt{{(X1-M)^2 + (X2-M)^2 ... + (Xna-M)^2}/Na}\)

SDb = \(\sqrt{{(Xna+1-M)^2 +... (Xnb-M)^2}/Nb}\)

SDaub^2 = \({(X1-M)^2 + (X2-M)^2 ... + (Xna-M)^2 + (Xna+1-M)^2 +... (Xnb-M)^2}/Na+Nb\)

SDaub^2 = \([ SDa^2*Na + SDb^2*Nb ] / [Na+Nb]\)

For Na = Nb = N

SDaub^2 = \([{SDa^2 + SDb^2}]N/2N\)

SDaub^2 = \([{SDa^2 + SDb^2}]/2\)

Hence, A is sufficient
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Set A and B are 2 sets of numbers. A has a standard 24 Sep 2010, 16:29
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great explanation. thank you
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Set A and B are 2 sets of numbers. A has a standard 27 Sep 2010, 14:02
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Dude man please don't post the answer so close to the question
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adishail
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Set A and B are 2 sets of numbers. A has a standard 27 Sep 2010, 14:28
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BeavisMan
Dude man please don't post the answer so close to the question
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It helps not to look down or scroll down while you are attempting to solve the question. :wink:
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utin
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Set A and B are 2 sets of numbers. A has a standard 01 Oct 2010, 05:49
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adishail
seekmba
Set A and B are 2 sets of numbers. A has a standard deviation 3 and a mean 5. B has the same mean but standard deviation 4. Can we find the standard deviation of the set A U B ? If no, then is it possible with the following
additional information?

(1) Both sets have the same number of members each.
(2) Both sets have distinct members (no number is common to both sets).

I found this on the forum here but did not understand why the OA is A. May be Bunuel can explain this.

Standard Dev for A = SDa

Standard Dev for B = SDb

Standard Dev for AUB = SDaub

Mean = M

Terms of A = X1, X2, ... Xna (Total terms Na)

Terms of B = Xna+1, Xna+2 , ... Xnb (Total terms Nb)

SDa = \(\sqrt{{(X1-M)^2 + (X2-M)^2 ... + (Xna-M)^2}/Na}\)

SDb = \(\sqrt{{(Xna+1-M)^2 +... (Xnb-M)^2}/Nb}\)

SDaub^2 = \({(X1-M)^2 + (X2-M)^2 ... + (Xna-M)^2 + (Xna+1-M)^2 +... (Xnb-M)^2}/Na+Nb\)

SDaub^2 = \([ SDa^2*Na + SDb^2*Nb ] / [Na+Nb]\)

For Na = Nb = N

SDaub^2 = \([{SDa^2 + SDb^2}]N/2N\)

SDaub^2 = \([{SDa^2 + SDb^2}]/2\)

Hence, A is sufficient
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Is it always the case that mean of the combined set will be equal to the mean of the individual sets.

I can't think of an example in which two sets have same no. of elements but each element must be different and still both the sets have same mean!!!

Please help...
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adishail
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Set A and B are 2 sets of numbers. A has a standard 04 Oct 2010, 08:22
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It is not necessary that the mean of a combined set is equal to the mean of the individual sets.

A = {2,2} => mean = 2

B = {5,5} => mean = 5

AUB = {2,2,5,5} => mean = 3.5
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Set A and B are 2 sets of numbers. A has a standard 16 Oct 2010, 07:19
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I always find these stats question to be tough man..
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