If sets A and B have the same number of terms, is the

Yes C is correct,

Standard deviation describes how the values in a set deviate from the mean.


(1) The range of set A is greater than the range of set B.
This gives only the range i.e. outermost values but what about the other values. consider -11,0,0,0,11 and -10,-10,0,10,10
now earlier one has higher range but lower standard deviation.

(2) Sets A and B are both evenly spaced sets.
Evently spaced is oK but what is spacing since that is required to know the deviation.

Both, equal no. of numbers and A has higher range means A has no. with greater spacing so A has more Standard deviation

Sets A and B have the same number of terms

Is SD of A > SD of B?

Statement 1

Range of A > Range of B

We know that they have the same number of terms but the range is not sufficient to determine the SD, the terms in between might have large or small deviations thus giving different measures of SD. Please refer to the above post for some examples

Statement 2

A,B are both evenly spaced. OK, so we know that both have the same number of terms and that they are both equally spaced, well they could even have the same components in which case the answer is NO.

Or one could be for example: 2,4,6,8 and the other 1,2,3,4 in which case the SD of the first one will be larger. Even in this case we don't know which one is A and which one is B, so this is clearly insufficient.

Both Statements together

We know that A must be larger since both are evenly spaced sets but the range of A is larger. Therefore, since they both have the same number of terms, only way that A can have a larger range is if components themselves are larger number. Therefore SD will always be larger.

Answer: C

Hope this helps
Cheers!
J

I request someone to explain the above problem along with concept in detail.

Thanks for the simple solution.
Kudos for you.

Good question.

Statement 1: very tempting to be sufficient, but not to be so.

In GMAT, always we have to try to break condition, as it is very easy to prove the condition

Say both A and B have 100 terms.
case 1:
A: {1,2,3,4......100}
B: {50,50,50..50} = SD = 0 , in this case, S.D of A > S.D of B

case 2(breaking case):
A: {1,2,3,4....100}
B:{1,98,98,98,98..98} => S.D of B is greater than that of A

two cases, insufficient

Statement 2: cleary insufficient, as we dont know which is more evenly spaced, is it A or B, accordingly bigger SD will be decided

1+2
given range of A > range of B, so A is more evenly spaced than B, so SD of A > SD of B => (C)

statement 1 alone is insufficient as there can be many scenario for example set A can be {1,2,9} and set B can be {11,12,13}
then standard deviation of set A is greater .
But when A is {1,1,3} and set B is {11,15,19} then it is insufficient .

Statement 2 is insufficient as it does not tell us anything about individual set

Together they are sufficient

Let us take two sets
Set A {1,3,5}
Set B {11,12,13}
Now range of set A is greater than that of set B
Mean is 3 of set A mean is 12 of set B .

While the ranges of two different sets can sometimes help to indicate their relative standard deviation, more information is needed in this case.

\(\# A = \# B\)

\({\sigma _A}\mathop > \limits^? {\sigma _B}\)

\(\left( 1 \right)\,\,{R_A} > {R_B}\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,{\rm{A = }}\left\{ {0,1} \right\}\,,\,\,B = \left\{ {0,0} \right\}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr \\
\,{\rm{Take}}\,\,\left\{ \matrix{\\
{\rm{A = }}\left\{ {0,0,0,10} \right\}\,\, \hfill \cr \\
B = \left\{ {0,0,9,9} \right\}\,\, \hfill \cr} \right.\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.\)

\(\left( 2 \right)\,\,{\rm{finite}}\,\,{\rm{APs}}\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,{\rm{A = }}\left\{ {0,2,4} \right\}\,,\,\,B = \left\{ {0,1,2} \right\}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr \\
\,{\rm{Take}}\,\,{\rm{A = }}\left\{ {0,1,2} \right\}\,,\,\,B = \left\{ {0,2,4} \right\}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,{\rm{distance}}\,\,{\rm{between}}\,\,{\rm{terms}}\,\,{\rm{and}}\,\,{\rm{mean}}\,\,{\rm{in}}\,\,A\,\,{\rm{is}}\,\,{\rm{larger}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Hi mikemcgarry

I read your blogs regarding standard deviation and statistics, both blogs were really amazing and insightful

Though, I couldnt apply the learning from those blogs in this question

Please help!!

Thanks

I don't think below red highlighted is correct.

Each statement alone is not sufficient.

Combined, we know the range of set A is greater than the range of set B. As well, both sets are evenly spaced.

Given these conditions, Set A will ALWAYS have a greater standard deviation.

For example:
Set A= {2, 4, 6, 8}
σ = 2.2

Set B = {2, 4, 6}
σ = 1.6

Answer is C.

Let us evaluate both the statements

Statement 1 The range of set A is greater than the range of set B.
Example 1: A= {1, 3, 5, 7, 10} ; B = {1, 2, 3, 4, 5}

Range of A = 10-1 = 9 ; Range of B = 5-1 = 4
Range of A is greater than Range of B; Also, since A has a higher spread, Standard Deviation of A is greater than Standard Deviation of B

Example 2:
A= {1,1,1, 13} ; B = {1,11,11,12}
Range of A > Range of B;
Standard Deviation of A = square root of ((1+1+1+169)/5) = square root of (172/5)
Standard Deviation of B = square root of ((1+121+121+144)/5) = square root of (387/5)

Therefore, Standard Deviation of B > Standard Deviation of A
Statement 1 is insufficient. (A) is rejected

Statement 2 Sets A and B are both evenly spaced sets.
Example 1: A= {100, 200, 300, 400} ; B= {1,2,3,4}
Both are evenly spaced. However, it is obvious that Standard Deviation of A > Standard Deviation of B

Example 2: A= {1,2,3,4} ; B= {100, 200, 300, 400} Basically, reverse of Example 1
Hence, Standard Deviation of B > Standard Deviation of A

Statement 2 in insufficient.
(B) is rejected

Combining 1 & 2: A and B have equal number of terms, Range of A is greater than Range of B, A & B are both equally spaced
Example 1: A= {100, 200, 300, 400} ; B= {1,2,3,4}
Certainly, Standard Deviation of A > Standard Deviation of B

Example 2: A = {100, 200, 300, 400} ; B= {1001, 1002, 1003, 1004}
Range of A = 300, Range of B= 3
You can observe that, each number in Set B in Example 2 is equal to 1000+ corresponding number in Set B in Example 1
Therefore, Standard Deviation of A > Standard Deviation of B
This stays in all the cases which satisfy both the conditions.

Therefore, (C) is the answer

MOOL MANTRA: If both the sets have evenly spaced terms, then set with longer range has larger Standard Deviation

Can someone please explain what is the meaning of evenly spaced sets?
How can we conclude that both Set A and Set B have same number of elements?

Official Explanation: While the ranges of two different sets can sometimes help to indicate their relative standard deviation, more information is needed in this case. Statement (1) tells us that the range of A is larger than the range of set B, but it does not guarantee that the standard deviation will be bigger. If Set A were (5, 10,10,10,10,15) it would have a range of 10 and a fairly small standard deviation as most terms are the same as the average. Set B could be the following (5,5,5,14,14,14) which would have a smaller range of 9 but a clearly larger standard deviation. Statement 1 is not sufficient as numerous scenarios are possible. Statement (2) is clearly insufficient by itself as nothing is known about the values in the sets. Taking the statements together, it is known that there are two evenly spaced sets with the same number of terms and A has a bigger range than B. This guarantees that A must have a higher dispersion around the mean and thus a higher standard deviation. The correct answer is C, both statements together are sufficient.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.