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Set A consists of five different numbers; set B consists of

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Set A consists of five different numbers; set B consists of

by sachin_yadav » Fri Jan 22, 2016 10:20 am
Set A consists of five different numbers; set B consists of four different numbers, each of which is in set A. Is the standard deviation of set A less than the standard deviation of set B ?

(1) Set A contains five consecutive integers.

(2) The average (arithmetic mean) of set A is equal to the average (arithmetic mean) of set B

OA is B
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by MartyMurray » Fri Jan 22, 2016 12:31 pm
The standard deviation is basically, thought not exactly, the average distance between each number in a set and the mean of the set.

So what this question is asking is basically whether the members of set A or set B are farther from their respective means.

Statement 1:

Let's look at a set of five consecutive integers as an example. Set A could be the following set.

{1, 2, 3, 4, 5}

The mean of this set is 3 and the average deviation from the mean is (2 + 1 + 0 + 1 + 2)/5 = 6/5. That's not exactly the standard deviation, but that will serve as an approximation.

Set B would contain only four of the integers from set A. Here is a possible version of set B.

{1, 2, 3, 4}

The standard deviation of this version should be less than that of set A, because the numbers in this set are on average closer to the mean of 2.5 than the numbers in set A are to 3.

Here is the average deviation. (1.5 + .5 + .5 + 1.5)/4 = 1

Since the average deviation of set A is 6/5 and that of set B is 1, without actually calculating their exact standard deviations I am going to guess that set B has a standard deviation smaller than that of set A.

Now look at this version of set B.

{1, 2, 4, 5}

Without even calculating you can see that without the 3 from set A, the numbers in this version of set B are on average further from the mean, 3, than are those in set A.

So set B could have a standard deviation smaller or greater than that of set A.

Insufficient.

Statement 2:

This tells us that even though set B has fewer elements is has the same mean as set A.

If all of the elements of set A and B are unique, then the only way that set B, with one fewer member, can have the same mean as set A is to have one member of A be equal to the mean of A and to create B by leaving that member out.

The deviation from the mean of the mean is 0. So as soon as we remove the member equal to the mean from A, we increase the average deviation from the mean of the elements. So if A and B share the same mean, B will always have a greater standard deviation.

Sufficient.

The correct answer is B.

(Note: My understanding is that on the GMAT the type of approximation I used in analyzing the value of Statement 1 will not be necessary. The way Statement 2 works, such that the standard deviation of one set is clearly greater than that of the other, is the type of situation you will be working with when answering official questions.)
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by Matt@VeritasPrep » Fri Jan 22, 2016 3:39 pm
Another way of thinking about this:

Given any set, the standard deviation will (more or less) go down if we add an EXPECTED number and go up if we add an UNEXPECTED number. For instance, if we take the set {1, 2, 3, 4} and put 2.5 in the set, we're adding an expected number: a number that's hardly unusual given the set we already had. But if we put 7 in the set, we're adding an unexpected number, something outside of the current range, so our standard deviation will go up.

S1:: Our set is something like {1, 2, 3, 4, 5}. So Set B consists of four of these. If B is {2, 3, 4, 5} it will have a SMALLER standard deviation, since A also includes an unexpected number (1). If B is {1, 2, 4, 5} it will have a LARGER standard deviation, since A also includes a very expected number (3). So this is NOT SUFFICIENT.

S2:: Suppose set A is {a, b, c, d, e}, not necessarily in increasing order (i.e. we could have a > c and c > b, or whatever). Suppose set B is {a, b, c, d}. Since the sets have the same mean, e must equal the average of set B. Hence set A is adding a very expected number (its own mean), causing its standard deviation to be less than B's; SUFFICIENT.
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by sachin_yadav » Sun Jan 24, 2016 10:17 am
Matt@VeritasPrep wrote:

S2:: Suppose set A is {a, b, c, d, e}, not necessarily in increasing order (i.e. we could have a > c and c > b, or whatever). Suppose set B is {a, b, c, d}. Since the sets have the same mean, e must equal the average of set B. Hence set A is adding a very expected number (its own mean), causing its standard deviation to be less than B's; SUFFICIENT.
Matt and Marty thank you for explanations.

Matt can you please elaborate more on your approach ?

Regards
Sachin
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