GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 17Live Online Classes with Top GRE Expert Rich Zwelling
08:00 PM EST
-10:00 PM EST
Get live instruction from a top-scoring, industry-recognized GRE teachers + 6 months of access to TTP Self-Study included! Tues/Thur Sept 7, 2026---Feb. 2, 2027, 1:00pm-3:00pm EST
Dec 14Live Online Classes with Top GRE Expert Michael Butville
08:00 PM EST
-10:00 PM EST
Get live instruction from a top-scoring, industry-recognized GRE teachers + 6 months of access to TTP Self-Study included! Mon/Wed Dec. 14, 2026---Feb. 22, 2027, 8:00pm-10:00pm EST
Standard Deviation of the given sets
[#permalink]
Updated on: 27 Feb 2020, 04:21
Updated on: 27 Feb 2020, 04:21
4
6
Bookmarks
00:00
Question Stats:
56% (00:39) correct
43% (00:29) wrong based on 48 sessions
Hide Show timer Statistics
Quantity A |
Quantity B |
The standard deviation of the set 10,20,30 |
The standard deviation of the set 10,20,20,20,20,20,30 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Can someone please explain me how the standard deviation can be determined in this case.
Both sets have same mean, same range, I just don't get the concept of more spread out.
Director
Joined: 16 May 2014
Posts: 590
Given Kudos: 0
GRE 1: Q165 V161
Re: Standard Deviation of the given sets
[#permalink]
22 Jun 2016, 01:33
22 Jun 2016, 01:33
5
Expert Reply
Definition
Lets understand the formula of Standard Deviation first (σ):
Variance = (SD)\(^2\) =σ\(^2\) = [∑\(( Xi - X )^2\) / N ]
Variance is the average of the squared differences from the mean, which basically a measure of how spread out numbers are.
So lets take the first case:
Case 1
i = 3
Numbers = 10, 20 , 30.
Xm = \(\frac{10 + 20 + 30}{3}\) = 20
Variance = \(\frac{(10-20)^2 + (20-20)^2 + (30-20)^2}{3}\) = 66.667
SD = σ = \(\sqrt{Variance}\) = 8.16
Case 2
i = 7
Numbers = 10, 20 , 20, 20 ,20 ,20 , 30.
Xm = 20 (Same as the previous one)
Variance = \(\frac{(10-20)^2 + (20-20)^2 + (20-20)^2 +(20-20)^2 +(20-20)^2 +(20-20)^2 +(30-20)^2}{7}\) = 28.57
SD = σ = \(\sqrt{Variance}\) = 5.34
So quantity A is greater.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Standard Deviation of the given sets
[#permalink]
24 Jun 2016, 11:06
24 Jun 2016, 11:06
6
3
Bookmarks
For the purposes of the GRE, it's sufficient to think of Standard Deviation as the Average Distance from the Mean. Here's what I mean:
Consider these two sets: Set A {7,9,10,14} and set B {1,8,13,18}
The mean of set A = 10 and the mean of set B = 10
How do the Standard Deviations compare? Well, since the numbers in set B deviate the more from the mean than do the numbers in set A, we can see that the standard deviation of set B must be greater than the standard deviation of set A.
Alternatively, let's examine the Average Distance from the Mean for each set.
Set A {7,9,10,14}
Mean = 10
7 is a distance of 3 from the mean of 10
9 is a distance of 1 from the mean of 10
10 is a distance of 0 from the mean of 10
14 is a distance of 4 from the mean of 10
So, the average distance from the mean = (3+1+0+4)/4 = 2
B {1,8,13,18}
Mean = 10
1 is a distance of 9 from the mean of 10
8 is a distance of 2 from the mean of 10
13 is a distance of 3 from the mean of 10
18 is a distance of 8 from the mean of 10
So, the average distance from the mean = (9+2+3+8)/4 = 5.5
IMPORTANT: I'm not saying that the Standard Deviation of set A equals 2, and I'm not saying that the Standard Deviation of set B equals 5.5 (They are reasonably close however).
What I am saying is that the average distance from the mean can help us see that the standard deviation of set B must be greater than the standard deviation of set A.
More importantly, the average distance from the mean is a useful way to think of standard deviation. This model is a convenient way to handle most standard deviation questions on the GRE.
More here:
Consider these two sets: Set A {7,9,10,14} and set B {1,8,13,18}
The mean of set A = 10 and the mean of set B = 10
How do the Standard Deviations compare? Well, since the numbers in set B deviate the more from the mean than do the numbers in set A, we can see that the standard deviation of set B must be greater than the standard deviation of set A.
Alternatively, let's examine the Average Distance from the Mean for each set.
Set A {7,9,10,14}
Mean = 10
7 is a distance of 3 from the mean of 10
9 is a distance of 1 from the mean of 10
10 is a distance of 0 from the mean of 10
14 is a distance of 4 from the mean of 10
So, the average distance from the mean = (3+1+0+4)/4 = 2
B {1,8,13,18}
Mean = 10
1 is a distance of 9 from the mean of 10
8 is a distance of 2 from the mean of 10
13 is a distance of 3 from the mean of 10
18 is a distance of 8 from the mean of 10
So, the average distance from the mean = (9+2+3+8)/4 = 5.5
IMPORTANT: I'm not saying that the Standard Deviation of set A equals 2, and I'm not saying that the Standard Deviation of set B equals 5.5 (They are reasonably close however).
What I am saying is that the average distance from the mean can help us see that the standard deviation of set B must be greater than the standard deviation of set A.
More importantly, the average distance from the mean is a useful way to think of standard deviation. This model is a convenient way to handle most standard deviation questions on the GRE.
More here:
