Range and SD

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Range and SD

by sampath » Fri Aug 12, 2011 9:45 pm
Set S and T have 100 numbers, respectively. Is standard deviation of S greater than or less than that of T?
(1) Range of S is greater that of T (2) Average of S is greater than that of T

OA given is E

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by GmatKiss » Sat Aug 13, 2011 2:59 am
[spoiler]IMO:E[/spoiler],

1)Range is the difference between highest and lowest value in a set.
say if we have 10 numbers in S and 90 numbers in T.
Range of S is 10 , and range of T is 9, But we cannot relate this info to determine the SD.

2)Average does not matter as we are not sure on the exact numbers in set S and T

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by veenu08 » Wed Jul 10, 2013 10:40 am
Can someone please explain this question. And it would be great if someone can let me know all the properties of SD that are important from the perspective of GMAT

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by Brent@GMATPrepNow » Thu Jul 11, 2013 7:07 am
sampath wrote:Set S and T have 100 numbers, respectively. Is standard deviation of S greater than or less than that of T?
(1) Range of S is greater that of T
(2) Average of S is greater than that of T
IMPORTANT: Rather than give examples with sets consisting of 100 numbers, let's make things easier and pretend that the question states that S and T have 4 numbers each. The same principles apply to sets consisting of 100 numbers.

Target question: Is standard deviation of S greater than or less than that of T?

Each statement alone is NOT SUFFICIENT, so let's jump straight to . . .

Statements 1 and 2 combined:
Let's examine two cases that comply with statements 1 and 2, yet yield conflicting answers to the target question.

Case a:
Set S = {-1, 0, 0, 1}. Range = 2, and Mean (average) = 0
Set T = {1, 1, 1, 1}. Range = 0, and Mean (average) = 1
In this case, the standard deviation of set S is greater than the standard deviation of set T

Case b:
Set S = {-100, 0, 0, 100}. Range = 200, and Mean (average) = 0
Set T = {-99, -99, 100, 100}. Range = 199, and Mean (average) = 0.5
In this case, the standard deviation of set S is less than the standard deviation of set T

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer = E

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by Uva@90 » Fri Jul 12, 2013 9:12 am
Brent@GMATPrepNow wrote:
sampath wrote:Set S and T have 100 numbers, respectively. Is standard deviation of S greater than or less than that of T?
(1) Range of S is greater that of T
(2) Average of S is greater than that of T
IMPORTANT: Rather than give examples with sets consisting of 100 numbers, let's make things easier and pretend that the question states that S and T have 4 numbers each. The same principles apply to sets consisting of 100 numbers.

Target question: Is standard deviation of S greater than or less than that of T?

Each statement alone is NOT SUFFICIENT, so let's jump straight to . . .

Statements 1 and 2 combined:
Let's examine two cases that comply with statements 1 and 2, yet yield conflicting answers to the target question.

Case a:
Set S = {-1, 0, 0, 1}. Range = 2, and Mean (average) = 0
Set T = {1, 1, 1, 1}. Range = 0, and Mean (average) = 1
In this case, the standard deviation of set S is greater than the standard deviation of set T

Case b:
Set S = {-100, 0, 0, 100}. Range = 200, and Mean (average) = 0
Set T = {-99, -99, 100, 100}. Range = 199, and Mean (average) = 0.5
In this case, the standard deviation of set S is less than the standard deviation of set T

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer = E

Cheers,
Brent
Brent, I have couple of doubts to be clarified.
Firstly, do we have to calculate the standard deviation on actual exam day. Don't you think it is really a time consuming one.
As you did
Statements 1 and 2 combined:
Let's examine two cases that comply with statements 1 and 2, yet yield conflicting answers to the target question.

Case a:
Set S = {-1, 0, 0, 1}. Range = 2, and Mean (average) = 0
Set T = {1, 1, 1, 1}. Range = 0, and Mean (average) = 1
In this case, the standard deviation of set S is greater than the standard deviation of set T

Case b:
Set S = {-100, 0, 0, 100}. Range = 200, and Mean (average) = 0
Set T = {-99, -99, 100, 100}. Range = 199, and Mean (average) = 0.5
In this case, the standard deviation of set S is less than the standard deviation of set T



Secondly, I have read in Kaplan GRE Book,
"How to deal with STANDARD DEVIATION
The Greater the spread, the higher the standard deviation"

So, as we go with above statement if greater the range, i.e. Spread, Greater should be the SD right?
If it is right then Statement 1 is sufficient isn't ?
Please correct me if I am wrong.

Regards,
Uva

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by Brent@GMATPrepNow » Fri Jul 12, 2013 9:35 am
Hi Uva,

We need not calculate the SD on the GMAT.
For the GMAT, it's typically a good idea to use the informal definition of SD, which is . . .
The Standard Deviation is approximately equal to the average distance the elements are away from the mean
Disclaimer: This is just an approximation. The definition does not give the actual SD. However, for the purposes of the GMAT, the informal definition is adequate for 99% of questions.


Okay, let's use the informal definition to examine the two cases:

Case a:
Set S = {-1, 0, 0, 1}. Range = 2, and Mean (average) = 0
If the mean is 0, then:
-1 is 1 away from the mean
0 is 0 away from the mean
0 is 0 away from the mean
1 is 1 away from the mean
So, the SD is roughly the average of 1, 0, 0 and 1. So the SD of set S is about 0.5

Set T = {1, 1, 1, 1}. Range = 0, and Mean (average) = 1
If the mean is 1, then:
1 is 0 away from the mean
1 is 0 away from the mean
1 is 0 away from the mean
1 is 0 away from the mean
So, the SD is the average of 0, 0, 0 and 0. So the SD of set S is 0
In this case, the standard deviation of set S is greater than the standard deviation of set T



Case b:
Set S = {-100, 0, 0, 100}. Range = 200, and Mean (average) = 0
If the mean is 0, then:
-100 is 100 away from the mean
0 is 0 away from the mean
0 is 0 away from the mean
100 is 100 away from the mean
So, the SD is roughly the average of 100, 0, 0 and 100. So the SD of set S is about 50


Set T = {-99, -99, 100, 100}. Range = 199, and Mean (average) = 0.5
If the mean is 0.5, then:
-99 is 99.5 away from the mean
-99 is 99.5 away from the mean
100 is 99.5 away from the mean
100 is 99.5 away from the mean
So, the SD is roughly the average of 99.5, 99.5, 99.5 and 99.5. So the SD of set S is about 99.5
In this case, the standard deviation of set S is less than the standard deviation of set T

I hope that helps.

Cheers,
Brent
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by Uva@90 » Fri Jul 12, 2013 10:09 am
Hi Brent,
Got your informal definition of SD. It really helps us to solve quickly.

Could you please clarify my second doubt ?

Secondly, I have read in Kaplan GRE Book,
"How to deal with STANDARD DEVIATION
The Greater the spread, the higher the standard deviation"

So, as we go with above statement if greater the range, i.e. Spread, Greater should be the SD right?
If it is right then Statement 1 is sufficient isn't ?
Please correct me if I am wrong.

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by Brent@GMATPrepNow » Fri Jul 12, 2013 11:32 am
Uva@90 wrote:Hi Brent,
Got your informal definition of SD. It really helps us to solve quickly.

Could you please clarify my second doubt ?

Secondly, I have read in Kaplan GRE Book,
"How to deal with STANDARD DEVIATION
The Greater the spread, the higher the standard deviation"

So, as we go with above statement if greater the range, i.e. Spread, Greater should be the SD right?
If it is right then Statement 1 is sufficient isn't ?
Please correct me if I am wrong.
"The Greater the spread, the higher the standard deviation"
This is a very general rule, and does not apply in every instance.

Many SD questions on the GMAT can be solved by "eyeballing" the numbers in the sets.

For example, a question might ask, "Which of the following sets has the greatest SD?"
A) {1, 51, 51, 51, 51, 51, 51, 51, 51, 101}
B) {1, 1, 51, 51, 51, 51, 51, 51, 101, 101}
C) {1, 1, 1, 1, 51, 51, 101, 101, 101, 101}
D) {{1, 1, 1, 1, 1, 101, 101, 101, 101, 101}
E) {101, 155, 155, 155, 155, 155, 155, 155, 155, 201}


Notice that the range is 100 for all 5 sets.
However, the answer here is D

Cheers,
Brent
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