A set consists of 20 different numbers whose average is 10.

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by [email protected] » Tue Dec 03, 2019 9:57 am
Hi All,

We're told that a set consists of 20 different numbers whose average is 10. We're asked which of the following pairs, when added to the set of numbers, must REDUCE the standard deviation of the set. This is an example of a 'concept' question; the GMAT will never expect you to actually calculate the Standard Deviation of a group of numbers, so as long as you know the concepts behind S.D., then you can answer this question without too much work.

Standard Deviation is essentially how 'spread out' a group of numbers is. In simple terms, the 'closer together' the group, the smaller the S.D. The 'farther apart' the group, the larger the S.D. Thus, to DECREASE the S.D. of a group of numbers, you need to include more numbers that are near the "middle" (re: the average) of the group. In this case, since the average is 10, including two more "10s" will decrease the S.D.

Final Answer: C

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by Brent@GMATPrepNow » Tue Dec 03, 2019 2:07 pm
Gmat_mission wrote:A set consists of 20 different numbers whose average is 10. Which of the following pairs, when added to the set of numbers, must reduce its standard deviation?

A. 0, 0
B. 0, 20
C. 10, 10
D. -10, 10
E. 5, 15

[spoiler]OA=C[/spoiler]

Source: Veritas Prep

-----ASIDE--------------------------------
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? 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 GMAT.
-----------------------------------------

Since all 20 numbers are DIFFERENT, we know that the standard deviation is greater than 0.

Since the mean of the given set is 10, adding two more 10's (both of which are zero units away from the mean) will DECREASE the standard deviation.

Answer: C

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Brent
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by Scott@TargetTestPrep » Thu Dec 05, 2019 7:01 pm
Gmat_mission wrote:A set consists of 20 different numbers whose average is 10. Which of the following pairs, when added to the set of numbers, must reduce its standard deviation?

A. 0, 0
B. 0, 20
C. 10, 10
D. -10, 10
E. 5, 15

[spoiler]OA=C[/spoiler]

Source: Veritas Prep
The standard deviation is a measure of how far the data values of a set are located from the average. When new values that are equal to the average are added to the set, the standard deviation is decreased because these new values have no deviation from the average.

In the given data set, since the average is 10, when 10 and 10 are added to the set, the standard deviation will decrease.

Answer: C

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