Is the standard deviation of a certain set greater than 5,000?

Explanation

The concept used in this question is that
Standard deviation of a certain list ≤ (1/2) range of that list

Statement 1) we do not have the information either about the terms in the list or about their range. So we cannot answer if the standard deviation of a certain set is greater than 5000.
So this statement is insufficient

Statement 2) we know that range is < 8000

Based on the relation between range and standard deviation i.e.
Standard deviation of a certain list ≤ (1/2) range of that list
We can conclude that

Standard deviation of this list < 4000

Though we do not know the exact standard deviation but still the information in statement is sufficient to conclude that
Standard deviation of this list ≤ 5000

Hence we have a definite NO as an answer for the question asked

Hence the statement (2) is sufficient

Answer is option B

The following is the reasoning behind the concept

Standard deviation of a certain set ≤ (1/2) range of the list

Reasoning

One of the convenient 5-step processes to calculate the numerical value of standard deviation is

1. Take the mean of the given terms
2. Take the difference of the terms from the mean
3. Take the Square of the differences
4. Take the average of the squares
5. Standard deviation = (average of the squares)1/2

Lets consider a list of two numbers (0, 10) and let’s calculate the standard deviation

1. Mean of the terms = (0 + 10)/2 = 5
2. On taking the difference of the terms from the mean we will get: 5, -5
3. On taking the Square of the differences we get: 25, 25
4. On taking the average of the squares we get: (25 + 25) / 2 = 25
5. So Standard deviation = (25)1/2
Standard deviation = 5

Conceptually we know that standard deviation signifies how deviated are the terms.
So more deviated are the terms the more is the standard deviation

So if there are only two terms the list will have maximum standard deviation, as the terms will be the extreme or in others maximum deviated.

So we see that in the above list the

Standard deviation = (1/2) range

So if we add a number between the two numbers in the above list then the terms in general will become closer to the mean and hence the standard deviation will become smaller.

For better clarity lets add another term 5 in the list witch had two numbers (0, 10)

Now the new list will become (0, 5,10) and lets calculate the standard deviation

1. Mean of the terms = (0 + 5 +10)/3 = 5
2. On taking the difference of the terms from the mean we will get: 5,0, -5
3. On taking the Square of the differences we get: 25, 0, 25
4. On taking the average of the squares we get: (25 +0 + 25) / 3 = 50/3
5. So Standard deviation = (50/3) 1/2





Now the key point to notice is that the proportion with which the denominator has increased in the 4 Th step, the numerator has not increased.

As in this case if we add a number between the original numbers then we can say the numerator will never increase with the same proportion as the denominator will increase.
Actually for the numerator to increase with the same proportion the 25 should have been added instead of ZERO, which will never be the case as we are adding a number between the previous range.

We can observe that
Standard deviation of the second list ≤ Standard deviation of the first list

Given: Standard deviation of the first list = (1/2) range of the first list

Hence we can say
Standard deviation of the second list ≤ (1/2) range of the first list

So we can conclude that

Standard deviation of a certain list ≤ (1/2) range of that list


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Alternate Reasoning

We can also say that standard deviation signifies the average distance of the terms from the mean.
Please note that this average is not same as Arithmetic Average. Technically, it is called Root-Mean-Squared Value, but for GMAT we believe you should KEEP IT SIMPLE, so let us not go in its details.

If we take two terms (0,10)
The average distance from the mean is 5.

If we add another term within 0 and 10 say 5 or 6 then the average distance of the new list will be LESS than the average distance of the terms in the previous list.

So we can say that

Standard deviation of the second list ≤ Standard deviation of the first list

Given: Standard deviation of the first list = (1/2) range of the first list

So we can conclude that

Standard deviation of a certain list ≤ (1/2) range of that list

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