Is the standard deviation of the numbers in list R less than the stand

Is the standard deviation of the numbers in list R less than the standard deviation of the numbers in list S ?

(1) The range of the numbers in R is less than the range of the numbers in S.
(2) Each number in R occurs once and each number in S is repeated.



nick1816 Thanks for your solution.

This problem has unusual wording, and I believe there is a misunderstanding of the word "repeated". I also read the question this way my first time too, and quickly chose B.

However, after slowing down and retrying and confirming with the OG, the word "repeated" does not mean ALL the same. It only means that each number is repeated more than once. As a result, we can pick values that make (2) insufficient.


Looking at both statements combined, with some example numbers, to see if we can get both a "Yes" and "No" answer:

Case A: "No", SD of R>S

R = {0,10}
S = {0,0,6,6,6,6,6,6,6,6,6,6,12,12} ---> most of the values are the same as the average of 6, so the standard deviation will be relatively low.


Case B: "Yes", SD of R<S

R = {10} ---> SD = 0
S = {0,0,12,12} ---> SD > 0

Insufficient together --> Answer is E


Note: We do not need the Standard Deviation formula used in the OG. We just need to know the concept that it measures how spread out the data is overall, relative to the mean. A few properties come up more often than others:
1) If we add the same value to each number in a set, the standard deviation doesn't change.
2) If we multiply each value by a constant, the standard deviation is multiplied by that constant.
3) If the range is 0, then the standard deviation is 0
4) As we see in statement 1, just because the range is lower doesn't mean that the standard deviation is lower. Range = Max - Min, whereas standard deviation measures the overall spread of all the values.
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Statement 1 -

Case 1 - S= {2,6}; R= {3,3,4,4,5,5}

SD of S >SD of R

Case 2- S= {2}; R={3,3,4,4,5,5}

SD of S< SD of R

Insufficient

Statement 2 -

Case 1 - S= {2,6}; R= {3,3,4,4,5,5}

SD of S >SD of R

Case 2- S= {2}; R={3,3,4,4,5,5}

SD of S< SD of R

Insufficient

E

GMATCoachBen

Thanks for the input. 'Each element is repeated' in the second statement actually got me. I've edited my solution.

BTW your solution is awesome. Kudos!

nick1816, thanks very much!

"repeated" got me too! It's a reminder of how humbling the GMAT can be; if we don't read with focus and precision, we pay for it! With my trainees, that's why I often emphasize the importance of great sleep and extremely focused training.
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In statement 1 if Range if R < S, does that not mean that the elements are farther away from the mean? Hence sufficient?


I got extremely confused with the input value process. Although I understood that in statement 2 , it is not necessary for call elementary of set s to be same.

Is there an easier way to get the correct answer

Posted from my mobile device

In your example, it looks like Case 2 does not satisfy statement 1.

Good question - the language of B is quite tricky!

KarishmaB Please help with this question.These questions from statistics are extremely hard.
Is standard deviation not dependent on range ?

Statement 1:
Suppose R has range 999, elements ranging from 1 to 1000
Suppose S has range 1000, elements ranging from 1 to 1001

We can't judge the standard deviation from the range. The standard deviation depends on how spread out (how scattered) the elements are.

Statement 2:
Statement 2 says that each number in S is repeated. It does not say that all the numbers are the same!

Again, the standard deviation depends on how spread out (how scattered) the elements are. We have no information on that.

Statements 1 and 2 together:
We still know nothing about how spread out the elements are.

So the answer is (E)

Posted from my mobile device
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KarishmaB

I would be so appreciative for your view on how to solve this problem. I did the following:

(1) The range of the numbers in R is less than the range of the numbers in S.
Range does not tell you about SD.

(2) Each number in R occurs once and each number in S is repeated.
Case 1:
For list S, you could have 1, 1, --> 0 SD
For list R, you could have: 1, 2 --> SD that is not 0

Case 2:
I was not sure what numbers to pick to disprove statement 2 with my second case

Is SD of R < SD of S?

How do we increase the SD of a set? By adding more numbers at the extremes (or removing from the middle)
How do we decrease the SD of a set? By adding more numbers at the middle (or removing from the extreme)

(1) The range of the numbers in R is less than the range of the numbers in S.

Normally, when range is greater, we might expect SD to be greater too
e.g. SD of {0, 1} < SD of {0, 100}

but that will not be the case always.
e.g. SD of {1, 100} > SD of {0, 50, 50, 50, 100}
because the second set has most elements at the mean.

(2) Each number in R occurs once and each number in S is repeated.

SD of {0, 100} > SD {0, 0, 1, 1}
SD of {0, 50, 100} = SD of {0, 0, 50, 50, 100, 100}
SD of {0, 50, 100} < SD of {0, 0, 0, 100, 100}
Not sufficient alone. There is no connection of SD when each number appears once vs when each number is repeated .

Using both,
SD of {0, 1} < SD of {0, 0, 100, 100}
SD of {1, 100} > SD of {0, 0, 50, 50, 50, 50, 50, 50, 100, 100}

Not sufficient.

Answer (E)
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KarishmaB

Thank you so much for your quick response. I understand your reasoning for statement 2, but for statement 1, both the second case you showed have the same range still? Do you have cases to show that the ranges are different per the statement? I am still having a bit of a difficult time connecting the dots unfortunately. Thank you again.

SD is a measure of how far apart the numbers are dispersed on the number line (from the mean).
Range is how far apart the minimum and the maximum are.

So one might think that when numbers have a greater range they will dispersed farther off from each other on the number line, but the point is that it may not be true.

For example,
e.g. SD of {0, 1} < SD of {0, 100}
SD of {2, 3} < SD of {20, 30}
Sets with lower range have lower SD.

But,
SD of {1, 100} > SD of {0, 50, 50, 50, 100}
Here, range of the LHS set is 99 while that of the RHS set is 100. But still the SD of the LHS is higher. RHS has many elements at the mean and hence its SD is lower.
So it is not necessary that a lower range will give lower SD.
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"Is the standard deviation of the numbers in list R less than the standard deviation of the numbers in list S ?"

the question itself says is stand dev of the number(S) in LIST R , then why should i take a case where there is one number in set R ?

Your explanation is the most helpful! I have a question about understanding sd in general.
For example, you said
SD of {1, 100} > SD of {0, 0, 50, 50, 50, 50, 50, 50, 100, 100}
How many 50s do we need to have here for the SD on the left to be larger? How many 50s do we need to have for the SDs to be equal? (if we keep number of 100 and 0s unchanged). For example, Is SD of {1,100} or SD of {0, 0, 50, 50, 100, 100} larger?
I am not sure when two SDs would be equal just by looking at the numbers.