Question: Comparing Standard Deviations

Comment on Comparing Standard Deviations

Hi Brent, I i show that Set A can comprise of X identical numbers (7,7,7....) and Set B can comprise of Y Identical numbers (11,11,11,11,11...) - then the SD will be 0 in both cases. would this example be enough to show the we cannot determine anything ? (Since logically, any other sets we choose will be different than 0)? Does this make sense?
gmat-admin's picture

That is sound reasoning :-)

I had choose 7 14 21. And 11 22 33 as my number and solved SD . Is the way okay
gmat-admin's picture

That's more work than is necessary, but it works (as long as you test TWO SETS of values for sets A and B).

Hello, I had issues with arriving at the correct answer for the question under 'Related Resources'. I used the 'Informal Definition' to compare the SD values and somehow two sets A and C seem to have same SD of 6. I couldn't figure out what could be wrong with my calcs
gmat-admin's picture

I'm sure your calculations are fine. The question isn't GMAT-worthy, because it requires us to apply the formal definition of SD, which is too tedious for the GMAT. I shouldn't have linked to it in the first place.

I have removed the link.

Just to clarify, in each of the following cases how would the standard deviation differ

Case 1:

[7, 14, 21]
[11, 22, 33]

Is the SD larger for the second case as the distance away from the mean (22) is greater?

Case 2

[7, 14, 21, 28]
[11, 22, 33]

What happens to the SD if the first set has got an extra multiple?
gmat-admin's picture

As I mention in the Standard Deviation video (https://www.gmatprepnow.com/module/gmat-statistics/video/806), the INFORMAL definition of Standard Deviation is sufficient for most questions on the GMAT.

Case I:
Set A: [7, 14, 21]
The mean = 14
7 is 7 units from the mean, 14 is 0 units from the mean, and 21 is 7 units from the mean.
Average distance from the mean = (7 + 0 + 7)/3 = 14/3

Set B: [11, 22, 33]
The mean = 22
11 is 11 units from the mean, 22 is 0 units from the mean, and 33 is 11 units from the mean.
Average distance from the mean = (11 + 0 + 11/3 = 22/3

22/3 > 14/3, so the Set B has a greater Standard Deviation
-------------------------

Case II
[7, 14, 21, 28]
The mean = 17.5
7 is 10.5 units from the mean, 14 is 3.5 units from the mean, 21 is 3.5 units from the mean, and 28 is 10.5 units from the mean.
Average distance from the mean = (10.5 + 3.5 + 3.5 + 10.5)/4 = 28/4 = 7

[11, 22, 33]
The mean = 22
11 is 11 units from the mean, 22 is 0 units from the mean, and 33 is 11 units from the mean.
Average distance from the mean = (11 + 0 + 11/3 = 22/3 = 7 1/3

In this case, the Standard Deviations (using the INFORMAL definition) are very close.
In fact, they're much too close to predict which one actually has the greatest Standard Deviation (using the FORMAL definition)

On the GMAT, you won't be asked to compare the Standard Deviations of set where the Standard Deviations are very close.

Cheers,
Brent

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