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The standard deviation of the dataset 10, 20, 30 [#permalink]
29 Jul 2018, 04:37
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Question Stats:
52% (00:26) correct
47% (00:27) wrong based on 69 sessions
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.
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Re: The standard deviation of the dataset 10, 20, 30 [#permalink]
05 Aug 2018, 14:26
We do not need to work this out till the very end. A: average= 20. xaverage= 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. xaverage= 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]
16 Oct 2019, 01:23



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Re: The standard deviation of the dataset 10, 20, 30 [#permalink]
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]
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





