There is a set A of 19 integers with mean 4 and standard dev

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There is a set A of 19 integers with mean 4 and standard deviation of 3. Now we form a new set B by adding 2 more elements to the set A. What two elements will decrease the standard deviation the most?
A) 9 and 3
B) -3 and 3
C) 6 and 1
D) 4 and 5
E) 5 and 5

OA is D

Why it can't be E ?
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by GMATGuruNY » Sat Jan 16, 2016 3:06 am
vinni.k wrote:There is a set A of 19 integers with mean 4 and standard deviation of 3. Now we form a new set B by adding 2 more elements to the set A. What two elements will decrease the standard deviation the most?
A) 9 and 3
B) -3 and 3
C) 6 and 1
D) 4 and 5
E) 5 and 5
In any given set, the CLOSER the data points are to the mean, the SMALLER the standard deviation.
Here, to yield the greatest decrease in the standard deviation, the two numbers added to the set must be AS CLOSE AS POSSIBLE to the mean of 4.

The correct answer is D.
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by Brent@GMATPrepNow » Sat Jan 16, 2016 8:21 am
Mitch's solution is perfect, so I won't rehash that.

Here's some additional info about Standard Deviation:

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.

Here are two free videos that cover everything you need to know:
- https://www.gmatprepnow.com/module/gmat- ... ics?id=806
- https://www.gmatprepnow.com/module/gmat- ... ics?id=809

Here are a few more practice questions where we can apply the concept of "average distance from the mean" as an approximation for Standard Deviation:

https://www.beatthegmat.com/standard-dev ... 74384.html
https://www.beatthegmat.com/standard-dev ... 69584.html
https://www.beatthegmat.com/range-and-sd-t89159.html


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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