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by Brent@GMATPrepNow » Fri May 29, 2015 1:05 pm
A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?
A). -1 and 9
B). 4 and 4
C). 3 and 5
D). 2 and 6
E). 0 and 8
For GMAT purposes, Standard Deviation (SD) can often be thought of as "the average distance the data points are away from the mean."

So, with {0, 2, 4, 6, 8}, the mean is 4.
0 is 4 units away from the mean of 4.
2 is 2 units away from the mean of 4.
4 is 0 units away from the mean of 4.
6 is 2 units away from the mean of 4.
8 is 4 units away from the mean of 4.

So, the SD can be thought of as the average of 4, 2, 0, 2, and 4. The average of these values is 2.4, so we'll say that the SD is about 2.4

Note: This, of course, isn't 100% accurate, but it's all you should really need for the GMAT.

Okay, so which pair of new numbers, when added to the original 5 numbers will yield a new SD that is closest to 2.4?

Well, to begin, it's useful to notice that each pair consists of numbers that are equidistant from the original mean of 4.
For example, in answer choice A, -1 is 5 units less than 4, and 6 is 5 units more than 4.
As such, add the two values in each answer choice will yield a mean of 4.

Okay, let's see what happens if we add -1 and 9 (answer choice A).
Well, -1 is 5 units away from the mean of 4, and 9 is 5 units away from the mean of 4. So, 5 and 5 will be added to 4, 2, 0, 2, and 4 to get a new SD. As you can see, this will result in a much larger SD.

Now, let's examine D (2 and 6)
Well, 2 is 2 units away from the mean of 4, and 6 is 2 units away from the mean. So, we'll be adding 2 and 2 to the five original differences of 4, 2, 0, 2, and 4. Since the average of 4, 2, 0, 2, and 4 is 2.4, adding differences of 2 and 2 should have the least effect on the original SD.

As such, the correct answer must be D

Cheers,
Brent
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by prachi18oct » Fri May 29, 2015 1:33 pm
Hi Brent,

Thanks for your explanation.
Although I find this question quite different from usual ones.Is there any other approach also ?
What level should this question be ?
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by Brent@GMATPrepNow » Fri May 29, 2015 1:40 pm
I'd say it's 700+ level. Here's a more formal approach:
A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?
A). -1 and 9
B). 4 and 4
C). 3 and 5
D). 2 and 6
E). 0 and 8
To begin, the SD formula looks like this:
Image

The set {0, 2, 4, 6, 8} has a mean of 4.
So, first we need to find the difference between each value and the mean.

0 is 4 units away from the mean of 4.
2 is 2 units away from the mean of 4.
4 is 0 units away from the mean of 4.
6 is 2 units away from the mean of 4.
8 is 4 units away from the mean of 4.

So, the SD will equal the square root of (4² + 2² + 0² + 2² + 4²)/5.
In other words, the SD = the square root of 40/5.
= the square root of 8

Okay, so which pair of new numbers, when added to the original 5 numbers will yield a new SD that is closest to the square root of 8?

Well, to begin, it's useful to notice that each pair consists of numbers that are equidistant from the original mean of 4.
For example, in answer choice A, -1 is 5 units less than 4, and 6 is 5 units more than 4.
As such, add the two values in each answer choice will yield a mean of 4.

Okay, let's see what happens if we add -1 and 9 (answer choice A).
Well, -1 is 5 units away from the mean of 4, and 9 is 5 units away from the mean of 4.
So, the new SD = the square root of (4² + 2² + 0² + 2² + 4² + 5² + 5²)/7.
= the square root of 90/7.
= the square root of approximately 13
This is considerably larger than the original SD of sqrt(8)

Now let's skip a few answers and try answer choice D.
Here, 2 is 2 units away from the mean of 4, and 6 is 2 units away from the mean.
So, the new SD = the square root of (4² + 2² + 0² + 2² + 4² + 2² + 2²)/7.
= the square root of 48/7.
= the square root of approximately 7
This one is pretty close to the original SD of sqrt(8).

In fact, if we check the other answer choices (lots of work!), we'll see that answer choice D is the best answer.

Cheers,
Brent
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by Brent@GMATPrepNow » Fri May 29, 2015 1:41 pm
I'd say this question is flirting on the edge of legitimacy (with regard to being a GMAT-worthy question).
The problem is that it takes too long to solve (unless there's a faster solution that I'm missing)

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by Ian Stewart » Fri May 29, 2015 3:34 pm
This actually is a real GMAT question, from one of the oldest versions of GMATPrep (though the original wording says "closest to" and not "close to"). It is, however, very different from almost every other standard deviation question I've seen. It's actually unlike almost every other GMAT question I've seen on any topic, because a proper mathematical proof here would take far too long. I gather we're meant to look at the original set and say "things in this set look like they're about 2 away from the mean", and then guess that the standard deviation won't change much if we add two new elements, each 2 away from the mean. I don't find that explanation all that satisfying though, because we don't get a very precise estimate of standard deviation just by eyeballing a set like this. The standard deviation here is actually closer to 3 than it is to 2.

Anyway, I wouldn't about this question. As long as you have a general idea of what standard deviation measures, you should be fine.
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by talaangoshtari » Sat May 30, 2015 12:06 am
If we have the set S, a set of distinct positive integers. The standarad deviation must increase in which of the following?
A. The smallest number is increased to become equal to the median

I assume that S={1, 2, 3}. For this set, mean is equal to 2, and the distance of each number from the mean is equal to -1, 0, 1, the mean is zero.

A. Since the smallest number should be equal to the median, and the median of set S is 2, then for A, S={2, 2, 3}. The mean is equal to 2.33 and the distance of each number from the mean is -.33, -.33, .67.So, the mean of these three numbers is equal to 0.01/3, which is larger than zero.

Based on this logic, I thought that this sentence must be true, but A is not correct.

Would you please provide an example that helps me understand why it is incorrect?
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by DavidG@VeritasPrep » Sat May 30, 2015 7:56 am
I assume that S={1, 2, 3}. For this set, mean is equal to 2, and the distance of each number from the mean is equal to -1, 0, 1, the mean is zero.
This is where you went awry - the distance from 1 to 2 is 1, not -1. (When calculating a standard deviation, the distance from each term to the mean will always be a nonnegative number.)

And you can think about this logically: the set of 1, 2, 3 is clearly more dispersed than the set of, say, 1, 1, 1.

The only time you'd have a standard deviation of 0 would be when every element of the set was exactly the same.
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by Amrabdelnaby » Thu Nov 19, 2015 6:14 am
Here the average distance is 2.4 or 2 5/7 of the mean so we should pic numbers that has the same average distance from the mean
A has an average of 5 away from the mean
B has an average of 0 away from the mean
C has an average of 1 away from the mean
D has an average of 2 away from the mean
E has an average of 4 away from the mean

So D has the closest average away fro mean to the SD so its the right answer.
Brent@GMATPrepNow wrote:I'd say it's 700+ level. Here's a more formal approach:
A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?
A). -1 and 9
B). 4 and 4
C). 3 and 5
D). 2 and 6
E). 0 and 8
To begin, the SD formula looks like this:
Image

The set {0, 2, 4, 6, 8} has a mean of 4.
So, first we need to find the difference between each value and the mean.

0 is 4 units away from the mean of 4.
2 is 2 units away from the mean of 4.
4 is 0 units away from the mean of 4.
6 is 2 units away from the mean of 4.
8 is 4 units away from the mean of 4.

So, the SD will equal the square root of (4² + 2² + 0² + 2² + 4²)/5.
In other words, the SD = the square root of 40/5.
= the square root of 8

Okay, so which pair of new numbers, when added to the original 5 numbers will yield a new SD that is closest to the square root of 8?

Well, to begin, it's useful to notice that each pair consists of numbers that are equidistant from the original mean of 4.
For example, in answer choice A, -1 is 5 units less than 4, and 6 is 5 units more than 4.
As such, add the two values in each answer choice will yield a mean of 4.

Okay, let's see what happens if we add -1 and 9 (answer choice A).
Well, -1 is 5 units away from the mean of 4, and 9 is 5 units away from the mean of 4.
So, the new SD = the square root of (4² + 2² + 0² + 2² + 4² + 5² + 5²)/7.
= the square root of 90/7.
= the square root of approximately 13
This is considerably larger than the original SD of sqrt(8)

Now let's skip a few answers and try answer choice D.
Here, 2 is 2 units away from the mean of 4, and 6 is 2 units away from the mean.
So, the new SD = the square root of (4² + 2² + 0² + 2² + 4² + 2² + 2²)/7.
= the square root of 48/7.
= the square root of approximately 7
This one is pretty close to the original SD of sqrt(8).

In fact, if we check the other answer choices (lots of work!), we'll see that answer choice D is the best answer.

Cheers,
Brent
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