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The standard deviation of the set 1, 5, 7, 19
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17 Jul 2019, 02:20
17 Jul 2019, 02:20
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Question Stats:
80% (00:34) correct
19% (00:23) wrong based on 76 sessions
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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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The standard deviation of the set 1, 5, 7, 19
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18 Nov 2019, 06:24
18 Nov 2019, 06:24
Carcass wrote:
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.
----ASIDE-----------------------
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 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.
-------------------------------
We have:
QUANTITY A: The standard deviation of the set 1, 5, 7, 19
QUANTITY B: The standard deviation of the set 0, 5, 7, 20
For this question, we should be able to "eyeball" the two sets to see that set B has a greater standard deviation.
Notice that the outer values of 0 and 20 make set B more dispersed than set A (which has outer values of 1 and 19.
As such the answer is B
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Re: The standard deviation of the set 1, 5, 7, 19
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06 Apr 2020, 13:50
06 Apr 2020, 13:50
wrong because i was very tired but it easy to see that 20 is further from the mean than 19 and 1 is closer to the mean than zero. Therefore B
