Standard Deviation more than S

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Standard Deviation more than S

by gmattesttaker2 » Sun Feb 23, 2014 4:35 pm
Hello,

Can you please tell me how to solve this:

Which of the following pairs of integers, when added to a list of 50 integers with a mean of 7
and standard deviation S will yield a list of integers with a standard deviation more than S?

(A) -8 and 8
(B) 6 and 7
(C) 7 and 7
(D) 8 and 8
(E) 8 and 9

OA: A

Thanks,
Sri
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by Brent@GMATPrepNow » Sun Feb 23, 2014 5:13 pm
gmattesttaker2 wrote:Hello,

Can you please tell me how to solve this:

Which of the following pairs of integers, when added to a list of 50 integers with a mean of 7
and standard deviation S will yield a list of integers with a standard deviation more than S?

(A) -8 and 8
(B) 6 and 7
(C) 7 and 7
(D) 8 and 8
(E) 8 and 9
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.
----------------------------

In the original question that you posted, the mean is 7.
When we check answer choice A, we see that -8 is a distance of 15 from the mean of 7 , and 8 is a distance of 1 from the mean of 7 .
When we check the other answer choices, we find that those other values are CLOSER to the mean of 7, than are the values in answer choice A

Cheers,
Brent
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by jain2016 » Wed Feb 10, 2016 10:25 am
In the original question that you posted, the mean is 7.
When we check answer choice A, we see that -8 is a distance of 15 from the mean of 7 , and 8 is a distance of 1 from the mean of 7 .
When we check the other answer choices, we find that those other values are CLOSER to the mean of 7, than are the values in answer choice A

Cheers,
Brent
[/quote]

Hi Brent ,

Can you please explain the above part.

Many thanks in advance.

SJ

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by MartyMurray » Wed Feb 10, 2016 4:07 pm
jain2016 wrote:
In the original question that you posted, the mean is 7.
When we check answer choice A, we see that -8 is a distance of 15 from the mean of 7 , and 8 is a distance of 1 from the mean of 7 .
When we check the other answer choices, we find that those other values are CLOSER to the mean of 7, than are the values in answer choice A

Cheers,
Brent
Hi Brent, Can you please explain the above part.
This question is not that well written, as we don't have any indication of the value of standard deviation S.

The mean is given as 7, but S could be any number 0 or greater.

So S could be 0, or S could be 10,000 or an even greater number.

The standard deviation of a set is basically the average distance between the values in the set and the mean of the set.

For instance, in set {2, 4, 6} the mean is 4. 2 and 6 are at a distance of 2 from 4, and 4 is 0 away from 4.

The mean of the set {0, 4, 8} is also 4, but 0 and 8 are farther from 4 than are 2 and 6. So this set has a standard deviation greater than that of the previous one.

If to the set {0, 4, 8}, were added -20 and 28, the mean would continue to be 4, but the new set, {-20, 0, 4, 8, 28} would have a standard deviation greater than that of the original set, as -20 and 28 are farther away from 4 than are 0 and 8, and so adding -20 and 28 to the set increases the average deviation from 4.

In the question being discussed here, we want to know which numbers would increase the standard deviation of the set. The truth is that without knowing the standard deviation of the set we can't tell whether ANY of those pairs would increase the standard deviation. So this question is actually unanswerable as written.

If the standard deviation, S, of the set without the additional pair is 0, then adding any of the pairs except 7 and 7 would increase the standard deviation.

If S of the set without the additional pair is 1000, then adding any of the pairs will reduce the standard deviation, because all of the numbers in the pairs are closer than 1000 away from 7. So adding them will reduce the average distance between the values and 7.

So what Brent sought to do is to just figure out which pair of numbers is on average farthest from 7. Even though we can't tell whether adding that pair will actually increase S, I guess that pair is the one most likely to increase S.

The pair that on average is farthest from 7 is the one in choice A, -8 and 8.
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by Matt@VeritasPrep » Thu Feb 11, 2016 5:33 pm
This is an awful question. Suppose my set is [350, 350, 350, ..., 350, 350, -336, -336, ..., -336, -336, -336]. Any of the options given will reduce the standard deviation!