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

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Intern
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A certain list of 50 data has mean of 6 and.... [#permalink]  10 Sep 2018, 22:37
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68% (00:33) correct 31% (01:21) wrong based on 32 sessions
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
Intern
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Re: A certain list of 50 data has mean of 6 and.... [#permalink]  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]  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]  14 Sep 2018, 11:59
Expert's post
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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Sandy
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Intern
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Re: A certain list of 50 data has mean of 6 and.... [#permalink]  02 Nov 2018, 23:30
sandy 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.
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Re: A certain list of 50 data has mean of 6 and.... [#permalink]  10 Jan 2019, 16:55
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sandy 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 $$x_i=x_{mean}$$.

0 and 12 definitely change the SD, but it will increase the SD because of the adding effect of value greater than SD.
_______________________________________________________________________
Please correct me if I am wrong.
Re: A certain list of 50 data has mean of 6 and....   [#permalink] 10 Jan 2019, 16:55
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# A certain list of 50 data has mean of 6 and....

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