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Standard Deviation of the given sets
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Updated on: 27 Feb 2020, 04:21
Updated on: 27 Feb 2020, 04:21
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Quantity A |
Quantity B |
The standard deviation of the set 10,20,30 |
The standard deviation of the set 10,20,20,20,20,20,30 |
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.
Can someone please explain me how the standard deviation can be determined in this case.
Both sets have same mean, same range, I just don't get the concept of more spread out.
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Joined: 16 May 2014
Posts: 590
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GRE 1: Q165 V161
Re: Standard Deviation of the given sets
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22 Jun 2016, 01:33
22 Jun 2016, 01:33
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Expert Reply
Definition
Lets understand the formula of Standard Deviation first (σ):
Variance = (SD)\(^2\) =σ\(^2\) = [∑\(( Xi - X )^2\) / N ]
Variance is the average of the squared differences from the mean, which basically a measure of how spread out numbers are.
So lets take the first case:
Case 1
i = 3
Numbers = 10, 20 , 30.
Xm = \(\frac{10 + 20 + 30}{3}\) = 20
Variance = \(\frac{(10-20)^2 + (20-20)^2 + (30-20)^2}{3}\) = 66.667
SD = σ = \(\sqrt{Variance}\) = 8.16
Case 2
i = 7
Numbers = 10, 20 , 20, 20 ,20 ,20 , 30.
Xm = 20 (Same as the previous one)
Variance = \(\frac{(10-20)^2 + (20-20)^2 + (20-20)^2 +(20-20)^2 +(20-20)^2 +(20-20)^2 +(30-20)^2}{7}\) = 28.57
SD = σ = \(\sqrt{Variance}\) = 5.34
So quantity A is greater.
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Re: Standard Deviation of the given sets
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24 Jun 2016, 11:06
24 Jun 2016, 11:06
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For the purposes of the GRE, it's sufficient to think of Standard Deviation as the Average Distance from the Mean. Here's what I mean:
Consider these two sets: Set A {7,9,10,14} and set B {1,8,13,18}
The mean of set A = 10 and the mean of set B = 10
How do the Standard Deviations compare? Well, since the numbers in set B deviate the more from the mean than do the numbers in set A, we can see that the standard deviation of set B must be greater than the standard deviation of set A.
Alternatively, let's examine the Average Distance from the Mean for each set.
Set A {7,9,10,14}
Mean = 10
7 is a distance of 3 from the mean of 10
9 is a distance of 1 from the mean of 10
10 is a distance of 0 from the mean of 10
14 is a distance of 4 from the mean of 10
So, the average distance from the mean = (3+1+0+4)/4 = 2
B {1,8,13,18}
Mean = 10
1 is a distance of 9 from the mean of 10
8 is a distance of 2 from the mean of 10
13 is a distance of 3 from the mean of 10
18 is a distance of 8 from the mean of 10
So, the average distance from the mean = (9+2+3+8)/4 = 5.5
IMPORTANT: I'm not saying that the Standard Deviation of set A equals 2, and I'm not saying that the Standard Deviation of set B equals 5.5 (They are reasonably close however).
What I am saying is that the average distance from the mean can help us see that the standard deviation of set B must be greater than the standard deviation of set A.
More importantly, the average distance from the mean is a useful way to think of standard deviation. This model is a convenient way to handle most standard deviation questions on the GRE.
More here:
Consider these two sets: Set A {7,9,10,14} and set B {1,8,13,18}
The mean of set A = 10 and the mean of set B = 10
How do the Standard Deviations compare? Well, since the numbers in set B deviate the more from the mean than do the numbers in set A, we can see that the standard deviation of set B must be greater than the standard deviation of set A.
Alternatively, let's examine the Average Distance from the Mean for each set.
Set A {7,9,10,14}
Mean = 10
7 is a distance of 3 from the mean of 10
9 is a distance of 1 from the mean of 10
10 is a distance of 0 from the mean of 10
14 is a distance of 4 from the mean of 10
So, the average distance from the mean = (3+1+0+4)/4 = 2
B {1,8,13,18}
Mean = 10
1 is a distance of 9 from the mean of 10
8 is a distance of 2 from the mean of 10
13 is a distance of 3 from the mean of 10
18 is a distance of 8 from the mean of 10
So, the average distance from the mean = (9+2+3+8)/4 = 5.5
IMPORTANT: I'm not saying that the Standard Deviation of set A equals 2, and I'm not saying that the Standard Deviation of set B equals 5.5 (They are reasonably close however).
What I am saying is that the average distance from the mean can help us see that the standard deviation of set B must be greater than the standard deviation of set A.
More importantly, the average distance from the mean is a useful way to think of standard deviation. This model is a convenient way to handle most standard deviation questions on the GRE.
More here:
