Given that set S has four odd integers and their range is 4, how many

Calculating all of the standard deviations is definitely unnecessary - you just have to have a proper understanding of the meaning of standard deviation.

{1,1,3,5} has an average of 2.5. The differences between the numbers and averages are {1.5,1.5,.5,2,5}
{1,3,5,5} has an average of 3.5. The differences between the numbers and averages are {2.5,.5,1.5,1.5}

Notice how the differences are the same - therefore the Standard Deviation will be the same. The same goes for the other two sets.

{1,1,1,5} has an average of 2. The differences between the numbers and averages are {1,1,1,3}
{1,5,5,5} has an average of 4. The differences between the numbers and averages are {3,1,1,1}

VERITAS PREP OFFICIAL SOLUTION:

Recall what standard deviation is. It measures the dispersion of all the elements from the mean. It doesn’t matter what the actual elements are and what the arithmetic mean is – the standard deviation of set {1, 3, 5} will be the same as the standard deviation of set {6, 8, 10} since in each set there are 3 elements such that one is at mean, one is 2 below the mean and one is 2 above the mean. So when we calculate the standard deviation, it will give us exactly the same value for both sets. Similarly, standard deviation of set {1, 3, 3, 5, 6} will be the same as standard deviation of {10, 12, 12, 14, 15} and so on. But note that the standard deviation of set {25, 27, 29, 29, 30} will be different because it represents a different arrangement on the number line.

Let’s look at the given question now.

Set S has four odd integers such that their range is 4. So it could look something like this {1, x, y, 5} when the elements are arranged in ascending order. Note that we have taken just one example of what set S could look like. There are innumerable other ways of representing it such as {3, x, y, 7} or {11, x, y, 15} etc.

Now in our example, x and y can take 3 different values: 1, 3 or 5

x and y could be same or different but x would always be smaller than or equal to y.
– If x and y were same, we could select the values of x and y in 3 different ways: both could be 1; both could be 3; both could be 5
– If x and y were different, we could select the values of x and y in 3C2 ways: x could be 1 and y could be 3; x could be 1 and y could be 5; x could be 3 and y could be 5.

For clarification, let’s enumerate the different ways in which we can write set S:
{1, 1, 1, 5}, {1, 3, 3, 5}, {1, 5, 5, 5}, {1, 1, 3, 5}, {1, 1, 5, 5}, {1, 3, 5, 5}

These are the 6 ways in which we can choose the numbers in our example.

Will all of them have unique standard deviations? Do all of them represent different distributions on the number line? Actually, no!

Standard deviations of {1, 1, 1, 5} and {1, 5, 5, 5} are the same. Why?

Standard deviation measures distance from mean. It has nothing to do with the actual value of mean and actual value of numbers. Note that the distribution of numbers on the number line is the same in both cases. The two sets are just mirror images of each other.

For the set {1, 1, 1, 5}, mean is 2. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean.

For the set {1, 5, 5, 5}, mean is 4. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean.

The deviations in both cases are the same -> 1, 1, 1 and 3. So when we square the deviations, add them up, divide by 4 and then find the square root, the figure we will get will be the same.

Similarly, {1, 1, 3, 5} and {1, 3, 5, 5} will have the same SD. Again, they are mirror images of each other on the number line.

The rest of the two sets: {1, 3, 3, 5} and {1, 1, 5, 5} will have distinct standard deviations since their distributions on the number line are unique.

In all, there are 4 different values that standard deviation can take in such a case.

Answer (B)

since range is 4 , then those 4 integers are close together and their must be a repeat among them and their could be 4 unique sets.

s = ( 2k+1, 2k+3,2k+5,2k+5), Mean = (8k+14)/4
s= (2k+1,2k+1,2k+3,2k+5), Mean = ( 8K+10)/4
s= (2k+1, 2k+3,2k+3, 2k+5), Mean = (8k+12)/4
s = (2k+1,2k+1,2k+1,2k+5) Mean = (8k+8)/4 ( this set's deviations are equivalent to S= (2K+1,2K+5,2K+5,2K+5,2K+5))

we ll get 4 different sets of deviations from the mean of each .... thus 4 different SD.......B

If the 4 odd integers in set S are positive and they are as small as possible, then S can be any one of the following sets:

{1, 1, 1, 5}

{1, 1, 3, 5}

{1, 1, 5, 5}

{1, 3, 3, 5}

{1, 3, 5, 5}

{1, 5, 5, 5}

We don’t have to consider any more sets since any other set will be equivalent to one of the sets above. For example, {7, 9, 9, 11} is equivalent to {1, 3, 3, 5} since the two sets will have the same standard deviation (SD). So we only need to consider the 6 sets above. However, of the 6 sets above, some of them actually have the same SD. For example, {1, 1, 1, 5} and {1, 5, 5, 5} will have the same SD, as will {1, 1, 3, 5} and {1, 3, 5, 5}. Therefore, there are only 4 distinct SDs (one each from {1, 1, 1, 5}, {1, 1, 3, 5}, {1, 1, 5, 5} and {1, 3, 3, 5}).

Answer: B

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.