A set of data consists of the following 5 numbers: 0, 2, 4

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

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This is a good explanation. You can also do it by looking at the numbers and using your common sense about standard deviation.

The average of the first list is 4 and the number deviate from 4 evenly. what i mean by that is, the next set of numbers (2,6) are both 2 away from 4, and the next set after that (0,8) are both 4 away from 4.

To have a similar deviation, we want the next set to look similar.

-1 and 9 will stretch the outter limits of the list, so that will increase the standard deviation significantly.
4 and 4 will add too much weight to the center of the list. It'll decrease the standard deviation.
3 and 5 may be right
2 and 6 may be right
0 and 8 will add weight to the outter limits again, and stretch the deviation.

So it's a toss up between (3 and 5) and (2 and 6). And my educated guess is on 2 and 6, since basically comes in right in the center of the previous standard deviation, it should change it the least.

For the record, I have never taught the standard deviation formula to any student of mine. I find it to be unnecessary for the GMAT when good, conceptual thinking can get you through.

D

From the given #s it is clear 4 is mean(average) of both sets.
If the variations of the sets are equal or close then their Stand Deviations are equal/close to each other too.

Variation of 1-set (5 #s) is 40/5=8
so Varition of set2 (7 #s) must be around 8 too, thus (40+ x)/7= around 8

x must be close to 16 so only D satisfies this value: (4-2)^2 +(4-6)^2=8
U can check others:
A) (4+1)^2+(4-9)^2=50
B) (4-4)^2+(4-4)^2=0
c) (4-3)^2+(4-5)^2=2
D) (4-2)^2 +(4-6)^2=8
E) (4-0)^2+(4-8)^2=32

if that is true then why not 3 and 5?

IMO it should be 4 and 4, as mean = new number, the change in deviation will be zero

yes, i missed one thing, to have minimum impact on SD, new values should be at equal distance from the mean and
if new numbers are more farther from the mean, less impact will be on SD.
if u look at the formula for SD, it will be more clear. if the values are same in that case in formula numerator will add 0 values for the the number(as mean will not change). however, denominator will increase by 2, thus the SD will be reduced considerably.
hope this will help

oh yea its 2 am here...lol...out of senses i guess and u also seems to be....

lol pls change numerator with denominator

PS; I got your point.

If the number are too far or too close to the mean the change will be bigger.
D is apparently not too far and too close to the means as compared to others.

The above method explains it well , and if you are short of time and need to make an educated guess , this works perfect.

If you are in for some calculations , this is how I got to it

mean = 4
sd = \sqrt{8} = 2.8


Expected values for the SD to not change are - One value below SD from mean is (4 - 2.8) = 1.2 , and one value above SD is (4 + 2.8) = 6.8
This would mean , adding 1.2 ans 6.8 would have no impact on the SD . SD remains the same when these two numbers are added. Now for SD to change the least , we need to add two values that are closest to these two values.

Hence any two values that are closest to 1.2 and 6.8 would change the SD , the least.

1. -1 , 9
distance between (1,9) and (1.2 and 6.8) is 2.2 and 2.2

2. 4 , 4
distance etween (4,4) and (1.2 , 6.8) is 2.8 and 2.8

3. 3 , 5
Distance is - 1.8 and 1.8

4. 2 , 6
Distance is - 0.8 and 0.8

5. 0 , 8
Distnace is - 1.2 and 1.2

Hence from above , we see that adding 2 and 6 , results in a value that would change the SD to the least. Hence D

-Jyothi

Given: {0,2,4,6,8}
Mean : 4
Let us not exact calculate S.D., we will stop one step before calculation
\(((0 - 4)^2 + (2-4)^2 + (4-4)^2 + (6-4)^2 + (8-4)^2)\) / 5 = 40/5 = 8

Let A = \((0 - 4)^2 + (2-4)^2 + (4-4)^2 + (6-4)^2 + (8-4)^2\)

Examining the options, options are given thankfully so that mean remains the same, after adding the options. (reducing our burden to calc mean again)

Option 1: -1 and 9
\(( A + (-1-4)^2 + (9-4)^2)/7 => (40 + 25 + 25) / 7\)= well over 8

Option 2: 4 and 4

\((A + (4-4)^2 + (4-4)^2)/7 = > 40/7\) = well below 8 = 5.x

Option 3: 3 and 5
\((A + (3-4)^2 + (5 - 4)^2)/7 => 42/7\) = 6

Option 4: 2 and 6
\((A + (2-4)^2 + (6-4)^2)/7 => 48/7\) = 6.9x (closer to 8)

Option 5: 0 and 8
\((A + (0-4)^2 + (8-4)^2)/7 => 72/7\) => 10.x (above 8)

option D

A very important question to understand Standard deviation. The 5 numbers on average are 2.4 points away from the mean...ie 12/5 =2.4.

Now we need to find a pair of 2 numbers such that their average deviation from mean is closest to 2.4. Or the sum of their deviation is closest to 4.8

Following are the average deviation in case of each option
A) -1, 9 --> Sum of deviation from mean(4) is equal to (5+5=10)
B) 4, 4 --> Sum of deviation from mean(4) is equal to (0+0=0)
C) 3,5 --> Sum of deviation from mean(4) is equal to (1+1=2)
D) 2,6 --> Sum of deviation from mean(4) is equal to (2+2=4) Closest to 4.8 . Hence, this is the correct answer
E) 0,8 --> Sum of deviation from mean(4) is equal to (4+4=8)

Hi guys, can you help me understand when to use N-1 (sample variance)/N (population variance) in the denominator of the variance equation
I see that in this question, using sample variance equation will give us answer E
While using population variance equation will give us answer D (OA)

Sample standard deviation of {0, 2, 4, 6, 8} (original set), with N - 1 in the denominator, is \(\sqrt{10}≈3.16\)
Sample standard deviation of {0, 2, 4, 6, 8, -1, 9} (option A), with N - 1 in the denominator, is \(\sqrt{5}≈3.9\)
Sample standard deviation of {0, 2, 4, 6, 8, 4, 4} (option B), with N - 1 in the denominator, is \(2\sqrt{\frac{5}{3}}≈2.58\)
Sample standard deviation of {0, 2, 4, 6, 8, 3, 5} (option C), with N - 1 in the denominator, is \(\sqrt{7}≈2.65\)
Sample standard deviation of {0, 2, 4, 6, 8, 2, 6} (option D), with N - 1 in the denominator, is \(2\sqrt{2} ≈ 2.83\)
Sample standard deviation of {0, 2, 4, 6, 8, 0, 8} (option E), with N -1 in the denominator, is \(2\sqrt{3} ≈ 3.46\)

B < C < D < Original < E < A. But E is closer to the original than D.

Population standard deviation of {0, 2, 4, 6, 8} (original set), with N in the denominator, is \(2\sqrt{2}≈2.83\)
Population standard deviation of {0, 2, 4, 6, 8, -1, 9} (option A), with N in the denominator, is \(3\sqrt{\frac{10}{7}}≈3.6\)
Population standard deviation of {0, 2, 4, 6, 8, 4, 4} (option B), with N in the denominator, is \(2\sqrt{\frac{10}{7}}≈2.39\)
Population standard deviation of {0, 2, 4, 6, 8, 3, 5} (option C), with N in the denominator, is \(\sqrt{6}≈2.44\)
Population standard deviation of {0, 2, 4, 6, 8, 2, 6} (option D), with N in the denominator, is \(4\sqrt{\frac{3}{7}}≈2.62\)
Population standard deviation of {0, 2, 4, 6, 8, 0, 8} (option E), with N in the denominator, is \(6\sqrt{\frac{2}{7}}≈3.2\)

B < C < D < Original < E < A. But D is closer to the original than E.

So, yes, you're correct. Depending on the formula used, you will get different answers. It's worth noting that "the standard deviation" on the GMAT refers to the population standard deviation. However, you you don't really need to know this and you won't need to calculate standard deviations on the GMAT. Instead, you need to understand the concept of standard deviation: GMAT questions on this topic are more conceptual than computational. The main thing you should know about standard deviation is that it measures the variation or dispersion of data points from the mean. A low standard deviation indicates that the data points are closely clustered around the mean, while a high standard deviation suggests that the data points are spread out over a larger range of values.


Therefore, the question above is NOT at all a realistic representation of what you would get on the GMAT, and it might not be worth your time to focus on it.
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Thanks a lot Bunuel