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The standard deviation of the set 1, 5, 7, 19 [#permalink]
03 Jul 2016, 16:03
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Quantity A 
Quantity B 
The standard deviation of the set 1, 5, 7, 19 
The standard deviation of the set 0, 5, 7, 20 
A)The quantity in Column A is greater. B)The quantity in Column 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 set 1, 5, 7, 19 [#permalink]
04 Jul 2016, 21:05
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Both sets have same mean, but second set has more dispersion from the mean, hence SD of the second set will be more. hence B



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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink]
01 Oct 2017, 06:14
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Just to be a little more formal I provide my solution.
Since the two groups have the same mean (= 10), we can get rid of the common numbers on the two lists, i.e. 5 and 7, since they would provide the same difference from the mean for the two columns. Then, we are left with 1 and 19 on column A and 0 and 20 on column B.
Instead of using the real standard deviation formula, we can get an idea using an approximate formula that is computed as the mean of the differences of the numbers from the mean. Thus, in column A we get 9+9/2 = 9 and on column B, 10+10/2 = 10. Thus, the standard deviation is higher in column B.
Answer B!



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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink]
01 Oct 2017, 07:36
IlCreatore wrote: Just to be a little more formal I provide my solution.
Since the two groups have the same mean (= 10), we can get rid of the common numbers on the two lists, i.e. 5 and 7, since they would provide the same difference from the mean for the two columns. Then, we are left with 1 and 19 on column A and 0 and 20 on column B.
Instead of using the real standard deviation formula, we can get an idea using an approximate formula that is computed as the mean of the differences of the numbers from the mean. Thus, in column A we get 9+9/2 = 9 and on column B, 10+10/2 = 10. Thus, the standard deviation is higher in column B.
Answer B! COuld you plz explain, i failed to understand 1. how you got the mean of setA = mean of set B = 10? 2. 9+9/2 = 9? how we get?
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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink]
01 Oct 2017, 07:47
pranab01 wrote: IlCreatore wrote: Just to be a little more formal I provide my solution.
Since the two groups have the same mean (= 10), we can get rid of the common numbers on the two lists, i.e. 5 and 7, since they would provide the same difference from the mean for the two columns. Then, we are left with 1 and 19 on column A and 0 and 20 on column B.
Instead of using the real standard deviation formula, we can get an idea using an approximate formula that is computed as the mean of the differences of the numbers from the mean. Thus, in column A we get 9+9/2 = 9 and on column B, 10+10/2 = 10. Thus, the standard deviation is higher in column B.
Answer B! COuld you plz explain, i failed to understand 1. how you got the mean of setA = mean of set B = 10? 2. 9+9/2 = 9? how we get? The results are that because I got rid of the numbers who are equal in the two columns, i.e. 5 and 7. In that way we get that the mean is 1+19/2 = 10 and 0+20/2 = 10. Than the sd are (101)+(1910)/2 = 9 and (100)(2010)/2 = 10. This is kind of an approximation that is faster to compute. However, you could have worked with the two full list as well. Getting the same mean of 8 for the two lists and sd = 5.5. for quantity A and sd = 6 for quantity B. Leading to the same answer, B!



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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink]
01 Oct 2017, 08:59
IlCreatore wrote: pranab01 wrote: IlCreatore wrote: Just to be a little more formal I provide my solution.
Since the two groups have the same mean (= 10), we can get rid of the common numbers on the two lists, i.e. 5 and 7, since they would provide the same difference from the mean for the two columns. Then, we are left with 1 and 19 on column A and 0 and 20 on column B.
Instead of using the real standard deviation formula, we can get an idea using an approximate formula that is computed as the mean of the differences of the numbers from the mean. Thus, in column A we get 9+9/2 = 9 and on column B, 10+10/2 = 10. Thus, the standard deviation is higher in column B.
Answer B! COuld you plz explain, i failed to understand 1. how you got the mean of setA = mean of set B = 10? 2. 9+9/2 = 9? how we get? The results are that because I got rid of the numbers who are equal in the two columns, i.e. 5 and 7. In that way we get that the mean is 1+19/2 = 10 and 0+20/2 = 10. Than the sd are (101)+(1910)/2 = 9 and (100)(2010)/2 = 10. This is kind of an approximation that is faster to compute. However, you could have worked with the two full list as well. Getting the same mean of 8 for the two lists and sd = 5.5. for quantity A and sd = 6 for quantity B. Leading to the same answer, B! From where have you got the formula for SD : Im not sure of that formula Secondly if we look at both qty's we can see that the smaller number in qty B < qty A and bigger number in qty B> qty A. So definitely QTY B is more spread out than QTY A. and so the option is B
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Re: The standard deviation of the set 1, 5, 7, 19
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