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M28-42

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New post 16 Sep 2014, 01:31
1
19
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A
B
C
D
E

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Re M28-42  [#permalink]

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New post 16 Sep 2014, 01:31
1
3
Official Solution:


(1) The standard deviation of set A is positive. We know that the standard deviation of any set is more than or equal to zero. The standard deviation of a set is zero only when the set consists of identical elements. So, this statement implies that set A does NOT consists of identical elements or that \(x\) does not equal to zero. Still this statement is not sufficient to answer the question.

(2) \(y=3\). The mean of set A is 3. Now, if \(x \neq 0\) for example if \(x=1\), then the standard deviation of B would be smaller than the standard deviation A, since the elements of B would be less widespread than the element of set A. But if \(x=0\), then \(A=\{3, \ 3, \ 3, \ 3, \ 3\}\) and \(B=\{3, \ 3, \ 3, \ 3, \ 3, \ 3\}\), so both will have the standard deviation of zero. Not sufficient.

(1)+(2) Since from (1) \(x \neq 0\), then adding a new element which equals to the mean will shrink the standard deviation, thus \(SD(A) \gt SD(B)\). Sufficient.


Answer: C
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Re: M28-42  [#permalink]

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New post 24 Jul 2017, 13:23
can someone please explain why mean of set A is 3?
thanks a ton
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Re: M28-42  [#permalink]

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New post 24 Jul 2017, 21:09
fitzpratik wrote:
can someone please explain why mean of set A is 3?
thanks a ton


I understood, As the terms are equally spaced out, Median = Mean. so as median is 3, Mean is 3. Am I right?
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New post 24 Jul 2017, 21:19
fitzpratik wrote:
I understood, As the terms are equally spaced out, Median = Mean. so as median is 3, Mean is 3. Am I right?


Yes, you're right. It's an equally spaced set (going up by x from one term to the next), so the median, which is 3, must also be the mean.
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Re: M28-42  [#permalink]

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New post 25 Jul 2017, 00:22
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Thanks Bunuel and IanStewart
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Re: M28-42  [#permalink]

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New post 03 Feb 2018, 01:53
Bunuel wrote:
Official Solution:


(1) The standard deviation of set A is positive. We know that the standard deviation of any set is more than or equal to zero. The standard deviation of a set is zero only when the set consists of identical elements. So, this statement implies that set A does NOT consists of identical elements or that \(x\) does not equal to zero. Still this statement is not sufficient to answer the question.

(2) \(y=3\). The mean of set A is 3. Now, if \(x \neq 0\) for example if \(x=1\), then the standard deviation of B would be smaller than the standard deviation A, since the elements of B would be less widespread than the element of set A. But if \(x=0\), then \(A=\{3, \ 3, \ 3, \ 3, \ 3\}\) and \(B=\{3, \ 3, \ 3, \ 3, \ 3, \ 3\}\), so both will have the standard deviation of zero. Not sufficient.

(1)+(2) Since from (1) \(x \neq 0\), then adding a new element which equals to the mean will shrink the standard deviation, thus \(SD(A) \gt SD(B)\). Sufficient.


Answer: C


Hi Bunuel,

Can you please elaborate statement 2 a little more? Even if x is 1 or higher, shouldn't the SD of the set B = the SD of set A as Y which is 3 is the same value as mean and the spread is pretty much the same. Thanks.
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Re: M28-42  [#permalink]

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New post 21 Apr 2018, 05:31
Hi Bunnel,


Sets A and B are equal, except for the term Y in set B.

Now, regardless of the value of X, if you don't consider the term Y in set B, the SD would be the same, correct? Because Set B = Set A

Now if we add the term 3 to Set B, the distance of Y to the average is 0, then the addition of the term to the set has no impact on the SD since the distance is zero. Thus, regardless of the value of X, the SD in both sets A and B are equal.

Could u please tell me where did I go wrong?


Thanks a lot
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Re: M28-42  [#permalink]

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New post 27 Aug 2018, 19:04
Below are the extracts from Important post, I made this same mistake, look below to revise the concept.


One thing we should know is the number of elements in the set, because the more elements we have the broader they are distributed relative to the mean.

Q: There is a set A of 19 integers with mean 4 and standard deviation of 3. Now we form a new set B by adding 2 more elements to the set A. What two elements will decrease the standard deviation the most?
A) 9 and 3
B) -3 and 3
C) 6 and 1
D) 4 and 5
E) 5 and 5
Solution: The closer to the mean, the greater decrease in standard deviation. D has 4 (equal our mean) and 5 (differs from mean only by 1). All other options have larger deviation from mean.
Eg:

Case 5: S = {1, 3, 5} or T = {1, 3, 3, 5}

The standard deviation (SD) of T will be less than the SD of S. Why? The mean of 1, 3 and 5 is 3. If you add another 3 to the list, the mean stays the same and the sum of the squared deviations is also the same but the number of elements increases. Hence, the SD decreases.
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Re: M28-42  [#permalink]

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New post 05 Dec 2018, 09:56
Bunuel wrote:
fitzpratik wrote:
can someone please explain why mean of set A is 3?
thanks a ton


\(A=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x\}\),

Mean = (the sum of the terms)/(the number of the terms) = (3 -2x + 3 - x + 3 + 3 + x + 3 + 2x)/5 = 15/5 = 3.

Hope it helps.


I corrected the divisor and result.
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Re: M28-42   [#permalink] 05 Dec 2018, 09:56
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