Set X consists of 100 numbers. The average (arithmetic mean) of set X is 10, and the standard deviation is 4.6. Which of the following two numbers, when added to set X, will decrease the set's standard deviation by the greatest amount?
A)-100 and -100
B)-10 and -10
C)0 and 0
D)0 and 20
E)10 and 10
OAE
Set X consists of 100 numbers. The average (arithmetic mean)
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Hi Gmatquant25,gmatquant25 wrote:Set X consists of 100 numbers. The average (arithmetic mean) of set X is 10, and the standard deviation is 4.6. Which of the following two numbers, when added to set X, will decrease the set's standard deviation by the greatest amount?
A)-100 and -100
B)-10 and -10
C)0 and 0
D)0 and 20
E)10 and 10
OAE
SD = Average of deviation from the mean.
Hear mean = 10.
Now we need to add two numbers such that average of deviation from the mean is less.
so if we add the mean itself then the deviation from the mean will be 0. Hence it reduces the average of deviation from the mean.
Hence answer is E
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Uva.
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Uva@90's solution is perfect, so I won't add to it.
I just wanted to elaborate on what Uva@90 said about "SD = Average of deviation from the mean"
For the purposes of the GMAT, 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?
One option is to recognize that, since the numbers in set B deviate the more from the mean than do the numbers in set A, 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 GMAT.
Cheers,
Brent
I just wanted to elaborate on what Uva@90 said about "SD = Average of deviation from the mean"
For the purposes of the GMAT, 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?
One option is to recognize that, since the numbers in set B deviate the more from the mean than do the numbers in set A, 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 GMAT.
Cheers,
Brent
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The standard deviation is a measure of the spread of data values around the mean. If data values are close to the mean, the standard deviation is small, and if data values are further away from the mean, the standard deviation is larger.gmatquant25 wrote:Set X consists of 100 numbers. The average (arithmetic mean) of set X is 10, and the standard deviation is 4.6. Which of the following two numbers, when added to set X, will decrease the set's standard deviation by the greatest amount?
A)-100 and -100
B)-10 and -10
C)0 and 0
D)0 and 20
E)10 and 10
Thus, in order to decrease the standard deviation, we want to find two values that are as close to the mean as possible. Since the mean is 10, the two values of 10 and 10 will decrease the standard deviation by the greatest amount.
Answer: E
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