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by yvonne12 » Tue Apr 10, 2007 3:26 pm
During an experiment, some water was removed from each of 6 water tanks. If the standard deviation of the volumes of water in the tanks at the beginning of the experiment was 10 gallons, what was the standard deviation of the volume of water in the tanks at the end of the experiment?

I. For each, 30 percent of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment.

II. The average volume of the water in the tanks at the end of the experiment was 63 gallons

can someone explain this to me please!!
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Source: — Data Sufficiency |
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by jayhawk2001 » Tue Apr 10, 2007 10:19 pm
Is the answer A ?

Let a, b, c, d, e and f be the volume of water in the 6 tanks.
Let x be the average

Std dev = 10 implies
sqrt ([ (a-x)^2 + (b-x)^2 + ... + (f-x)^2] / 6) = 10

1 - sufficient. Knowing that 30% was removed from each tank tells
us that a becomes 0.7a, b becomes 0.7b etc. At the same time
x, the average becomes 0.7x

So, new standard deviation is
sqrt ([ (0.7a-0.7x)^2 + (0.7b-0.7x)^2 + ... + (0.7f-0.7x)^2] / 6)
which is 0.7 * old standard deviation

2 - insufficient. Knowing the average after the removal doesn't give
us any relationship between x and a, b, c, d, e, f.
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right it is A

by yvonne12 » Wed Apr 11, 2007 4:11 pm
thanks, I guess I dont have a thorough understanding of this topic. Im going to study it.
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by aj5105 » Mon Jan 12, 2009 8:31 pm
Lecture on Mean, Variance & SD

https://video.google.com/videoplay?docid ... 0168799472
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by logitech » Mon Jan 12, 2009 11:02 pm
aj5105 wrote:Lecture on Mean, Variance & SD

https://video.google.com/videoplay?docid ... 0168799472
EVERYBODY SHOULD WATCH THIS!

Thanks Aj5105!
LGTCH
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by welcome » Tue Jan 13, 2009 8:27 am
Very good link, Thanks aj5105.
Shubham.
590 >> 630 >> 640 >> 610 >> 600 >> 640 >> 590 >> 640 >> 590 >> 590
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by California4jx » Sun May 17, 2009 11:46 am
aj5105 wrote:Lecture on Mean, Variance & SD

https://video.google.com/videoplay?docid ... 0168799472
very good explanation in that link -- thanks aj !
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by abcdefg » Mon Jul 20, 2009 12:42 pm
so to summarize, the standard deviation will not change if:
1) You add or subtract a constant to each term
2) Increase or decrease each term in a set of terms by the same percentage

correct?
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by Domnu » Mon Jul 20, 2009 12:55 pm
Here's why... if you multiply everything by a constant .7 (which decreases all terms by 30%), you end up calculating

((.7*a1-.7*u)^2 + (.7*a2-.7*u)^2 + ... + (.7*a6-.7*u)^2)/6

which can be expressed in terms of the original standard deviation.
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by real2008 » Mon Jul 20, 2009 12:58 pm
Domnu wrote:Here's why... if you multiply everything by a constant .7 (which decreases all terms by 30%), you end up calculating

((.7*a1-.7*u)^2 + (.7*a2-.7*u)^2 + ... + (.7*a6-.7*u)^2)/6

which can be expressed in terms of the original standard deviation.
so i suppose D is the answer!
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by Domnu » Mon Jul 20, 2009 1:03 pm
real2008 wrote:
Domnu wrote:Here's why... if you multiply everything by a constant .7 (which decreases all terms by 30%), you end up calculating

((.7*a1-.7*u)^2 + (.7*a2-.7*u)^2 + ... + (.7*a6-.7*u)^2)/6

which can be expressed in terms of the original standard deviation.
so i suppose D is the answer!
Actually, the answer should be A. The fact that the average at the end of the experiment was 63 doesn't change anything.
Have you wondered how you could have found such a treasure? -T
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by real2008 » Mon Jul 20, 2009 9:12 pm
Domnu wrote:
real2008 wrote:
Domnu wrote:Here's why... if you multiply everything by a constant .7 (which decreases all terms by 30%), you end up calculating

((.7*a1-.7*u)^2 + (.7*a2-.7*u)^2 + ... + (.7*a6-.7*u)^2)/6

which can be expressed in terms of the original standard deviation.
so i suppose D is the answer!
Actually, the answer should be A. The fact that the average at the end of the experiment was 63 doesn't change anything.
I got it. Thank you.
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