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The standard deviation of the dataset 10, 20, 30

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The standard deviation of the dataset 10, 20, 30 [#permalink] New post 29 Jul 2018, 04:37
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Quantity A
Quantity B
The standard deviation of the dataset 10, 20, 30
The standard deviation of the dataset 10, 20, 20, 20, 20, 20, 30


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: The standard deviation of the dataset 10, 20, 30 [#permalink] New post 05 Aug 2018, 14:26
We do not need to work this out till the very end.
A: average= 20. x-average= -10,0,10 respectively. Square the differences- 100, 0, 100. Average of the squared differences- 200/3. Standard deviation= sqroot (200/3)
B: average=20. x-average= -10, 0,0,0,0,0, 10 respectively. Square the differences- 100, 0,0,0,0,0,100. Average of squared differences- 200/7. Standard deviation= sqroot(200/7).

Since denominator of A is smaller than that of B, A will be the larger value.

I also understand this is a very time consuming, tedious, long method. So if someone has a quicker way, please help
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Re: The standard deviation of the dataset 10, 20, 30 [#permalink] New post 16 Oct 2019, 01:23
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Re: The standard deviation of the dataset 10, 20, 30 [#permalink] New post 25 Feb 2020, 22:47
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Can't we simply eyeball the standard deviation here?

Standard deviation measures 'how far' a set of values are from the arithmetic mean of that set, and we know µa and µb=20. Looking at B we know that there are 5 values that correspond with the µ so the standard deviation will be smaller as the spread will be smaller, meaning that we can safely say SD A> SD B. Simply put, there is more variability in A.
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Re: The standard deviation of the dataset 10, 20, 30 [#permalink] New post 06 Apr 2020, 13:27
I went for the eyeball method with a calculation of the mean for quantity B just to be sure it was 20 (probably a waste of time)
Re: The standard deviation of the dataset 10, 20, 30   [#permalink] 06 Apr 2020, 13:27
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