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# M28-42

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16 Sep 2014, 00:31
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Question Stats:

40% (01:20) correct 60% (00:56) wrong based on 43 sessions

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Set $$A=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x\}$$, where $$x$$ is an integer. Is the standard deviation of set A more than the standard deviation of set $$B=\{3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x, \ y\}$$

(1) The standard deviation of set A is positive.

(2) $$y=3$$.
[Reveal] Spoiler: OA

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16 Sep 2014, 00:31
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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.

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24 Jul 2017, 12:23
can someone please explain why mean of set A is 3?
thanks a ton
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24 Jul 2017, 20:04
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 = 15/3 = 5.

Hope it helps.
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24 Jul 2017, 20:09
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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24 Jul 2017, 20:19
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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24 Jul 2017, 23:22
Thanks Bunuel and IanStewart
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03 Feb 2018, 00:53
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.

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] 03 Feb 2018, 00:53
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# M28-42

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