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Re: Set A consists of four distinct numbers; set B consists of f [#permalink]
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Hello,

First of all, let's assume set A = {1,2,3,4}
Mean of Set A= ( 1 + 2 + 3 + 4 ) / 4 = 2.5
Since set B has all values of Set A, Set B can be { 1, 2, 3, 4, 2.5 }
Mean of Set B = ( 1 + 2 + 3 + 4 + 2.5) / 5 = 2.5
Calculate the variance for both sets independently
For Set A,
Sum (x^2) = 30
The variance of A = 1.67

For Set B,
Sum (x^2) = 36.25
Variance of B = 1.25
After finding the variance, find Standard deviation using sqrt (variance).
SD of A = 1.291
SD of B = 1.25

After solving, it is clear that the variation of A is more than the variation of B. This is because the values in set a are less than the number of values in b
Quantity A is greater
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Re: Set A consists of four distinct numbers; set B consists of f [#permalink]
What if we choose -2,-1,0,1,2 for B and -2,-1,1,2 for A, would not the S.D of these equal to each other.
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Re: Set A consists of four distinct numbers; set B consists of f [#permalink]
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basit wrote:
What if we choose -2,-1,0,1,2 for B and -2,-1,1,2 for A, would not the S.D of these equal to each other.


No in the first case would be two and in the second case would be one

They spread apart more

More on SD here https://gre.myprepclub.com/forum/gre-mean- ... tml#p83258
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Set A consists of four distinct numbers; set B consists of f [#permalink]
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Regardless of what numbers we choose for set A, in order for set B to have the same mean, the one extra value in set B must equal the mean of set A.

Given that deviation can be roughly described as a measure of average distance from the mean, set B will have four values that are the same distance from the mean as we see in set A, plus one value that equals the mean (distance from mean = 0), so any kind of average measure of distance from the mean will be less in set B.
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Re: Set A consists of four distinct numbers; set B consists of f [#permalink]
It seldom happens that you are required to calculate the SD.
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Re: Set A consists of four distinct numbers; set B consists of f [#permalink]
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Yup correct, No need to calculate the SD accurately. The new value added to set B is the mean of set A.
The sum of squares of differences for respective values for both the sets will be the same since the only value added for set B is (mean - mean)^2=0.
But the no of values is 4 for set A and 5 for set B which is actually the denominator in the SD formula.
Since the denominator of SD for set A < the denominator of SD for set B therefore SD(A) > SD(B).
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