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A certain list of 50 data has mean of 6 and....

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A certain list of 50 data has mean of 6 and.... [#permalink] New post 10 Sep 2018, 22:37
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A certain list of 50 data has mean of 6 and a standard deviation of D, where D is positive.Which of the following pairs of the data, when added to the list, must result in a list of 52 data with standard deviation less than D?


A)-6 and 0


B)0 and 0


C)0 and 6


D)0 and 12


E)6 and 6
[Reveal] Spoiler: OA
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Re: A certain list of 50 data has mean of 6 and.... [#permalink] New post 12 Sep 2018, 00:38
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The general rule is (Magoosh):
If you do anything that “bunches the numbers together”, that decrease the standard deviation. If you do anything that “pulls the numbers further apart”, that increase the standard deviation.

And also:
In order to decrease the standard deviation of a set of numbers, you must add a value that is less than one standard deviation away from the mean

Given that standard deviation is positive (meaning, it is not equal to zero), then it means that adding 2 numbers with zero standard deviation will decrease the standard deviation of the set.

E) stand.dev of 6 and 6 is zero. It must decrease stand.dev
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Re: A certain list of 50 data has mean of 6 and.... [#permalink] New post 13 Sep 2018, 00:45
I Picked D --> 0 and 12. why won't the SD change?
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Re: A certain list of 50 data has mean of 6 and.... [#permalink] New post 14 Sep 2018, 11:59
Expert's post
Madhavi1990 wrote:
I Picked D --> 0 and 12. why won't the SD change?


Any number other than the mean would change the SD.

Why?

We need to look at the formula for standard deveation for this.

\(s=\sqrt{\frac{(x_i - x_{mean})^2}{N-1}}\). Now if you pick numbers and 12 still \((x_i - x_{mean})^2\) is a positive number in both cases.

The ONLY time the new numbers have no effect on standard deveation is when \(x_i=x_{mean}\).
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Re: A certain list of 50 data has mean of 6 and.... [#permalink] New post 02 Nov 2018, 23:30
sandy wrote:
Madhavi1990 wrote:
I Picked D --> 0 and 12. why won't the SD change?


Any number other than the mean would change the SD.

Why?

We need to look at the formula for standard deveation for this.

\(s=\sqrt{\frac{(x_i - x_{mean})^2}{N-1}}\). Now if you pick numbers and 12 still \((x_i - x_{mean})^2\) is a positive number in both cases.

The ONLY time the new numbers have no effect on standard deveation is when \(x_i=x_{mean}\).



But here, this is decreasing SD and not keeping it same.
Re: A certain list of 50 data has mean of 6 and....   [#permalink] 02 Nov 2018, 23:30
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