standard deviation

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standard deviation

by baller12q » Sun Mar 29, 2009 12:18 pm
a certain list of 100 data has an average (arithmetic mean) of 6 and a standard deviation of D, where D is positive. Which of the following pairs of data, when added to the list must result in a list of 102 data with standard deviation less than D.

a. -6 & 0
b. 0 & 0
c. 0 & 6
d. 0 & 12
e. 6 & 6
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by vittalgmat » Sun Mar 29, 2009 6:19 pm
IMO it is E.

Since the mean is 6, Adding any value that is = mean will reduce the deviation.


ht helps
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by manoj0609 » Tue Mar 31, 2009 2:43 pm
Definetly the ans should be E .

SD is more when the difference between mean and a datapoint is large
and viceversa.

here we have to reduce the SD which means need to include datapoints in such a way that difference is as low as possible. if we add 6 and 6 the diff betn mean and these poitns is zero i.e least possible and hence this can give u least SD among all the options.

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by dtweah » Sun Apr 05, 2009 4:50 am
manoj0609 wrote:Definetly the ans should be E .

SD is more when the difference between mean and a datapoint is large
and viceversa.

here we have to reduce the SD which means need to include datapoints in such a way that difference is as low as possible. if we add 6 and 6 the diff betn mean and these poitns is zero i.e least possible and hence this can give u least SD among all the options.

Cheers,manu
The reasoning for the answer being E is not b/c the difference between the added data points and the mean is the least possible. Clearly the difference between -6 and 6 (-6-6=-12 <6-6=0) is smaller. The key is in the calculation of standard deviation,THESE DIFFERENCES ARE SQUARED so that the least becomes the greatest IF THE LEAST IS NEGATIVE!!. (-12)^2 >0^2. If the answer choice did not have a negative data point, your reasoning would have been valid, since O, which is neither positive nor negative is the least of all positive numbers squared. See this as a CR exercise which though you reasoned wrongly you still got right. (laugh),

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by manoj0609 » Sun Apr 05, 2009 5:37 am
Could you please elaborate more on this? I am not quite convinced with your reasoning.

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by dtweah » Sun Apr 05, 2009 10:45 am
manoj0609 wrote:Could you please elaborate more on this? I am not quite convinced with your reasoning.
Original Data

1, 1

Mean 1+1/2 = 1

Standard Deviation (((1-1)^2+(1-1)^2))/2)^.5 =0

Case I: Additional data: 1

New data 1, 1, 1

Mean 1+1+1/3=1


SD: (((1-1)^2+(1-1)^2)+(1-1)^2))/3)^.5 =0

SD 0

Case II Additional Data -1

new data 1, 1, -1

Mean 1+1-1/3 =1/3

SD:( ((1- 1/3)^2 +(1- 1/3)^2) + (-1-1/3)^2)/3)^.5

Now the point I am making can be seen in the third parenthesis above. -1-1/3 is the least of corresponding expressions without squaring them. But you cannot say so when we square them.

((2/3)^2 + (2/3)^2 + (4/3)^2)/3)^.5

(4/3)^2 > (2/3)^2

SD in this case = 1.632
It is because this squared expression adds a bigger term in the numerator of the SD fraction that the SD shoots up higher above the previous cases, though without it the SD would still have been higher, since 1-1/3 is different from 0 ( se the other cases) and the squre of any number different from zero is greater than zero. IF ANYTHING THIS IS THE ONLY REASON WHY THE SD GOES UP, NOT EVEN BECAUSE OF THE SQUARED NEGATIVE TERM, SINCE THAT ONLY MAKES IT HIGHER.

I am only getting at the bottom of the LEAST in your reasoning, ie that least when squred becomes greatest. Which is the cause of a bigger increase, the Lease or the Greatest? I go with the Greatest because there is no least after squaring. If you stay with the Least, you may not calculate the standard deviation since that requires squaring which turns least into greatest. Hope this is clear.

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by manoj0609 » Sun Apr 05, 2009 11:26 am
Thanks mate !

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by Ian Stewart » Sun Apr 05, 2009 5:02 pm
dtweah wrote: The reasoning for the answer being E is not b/c the difference between the added data points and the mean is the least possible.
manu's reasoning above was perfectly valid. Perhaps it would be more clear to say 'positive difference', or 'distance', rather than simply 'difference', but the logic was otherwise fine.

In general, when thinking about standard deviation, you should think in terms of distances to the mean, and not differences from the mean; otherwise you might be distracted by negative signs that aren't at all important. That is, if you're looking at this set: {1, 2, 3, 4, 5}, if you calculate the difference ("mean - x") for each x in the set, some of these differences will be positive, some negative. Those negative signs aren't important, since standard deviation is only based on distances from each element to the average - you can feed the numbers 2, 1, 0, 1, 2 into the standard deviation calculation to get the correct answer. Mind you, I've yet to see a real GMAT question that requires you to calculate standard deviation.
dtweah wrote:
SD:( ((1- 1/3)^2 +(1- 1/3)^2) + (-1-1/3)^2)/3)^.5

Now the point I am making can be seen in the third parenthesis above. -1-1/3 is the least of corresponding expressions without squaring them. But you cannot say so when we square them.

((2/3)^2 + (2/3)^2 + (4/3)^2)/3)^.5

(4/3)^2 > (2/3)^2

SD in this case = 1.632
Not sure how you got 1.632 here. The standard deviation of {-1, 1, 1} is:

sqrt[(4/9 + 4/9 + 16/9)/3] = sqrt(8/9) = [2*sqrt(2)]/3, which is certainly less than 1.
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by dtweah » Sun Apr 05, 2009 5:21 pm
Ian Stewart wrote:
dtweah wrote: The reasoning for the answer being E is not b/c the difference between the added data points and the mean is the least possible.
manu's reasoning above was perfectly valid. Perhaps it would be more clear to say 'positive difference', or 'distance', rather than simply 'difference', but the logic was otherwise fine.

In general, when thinking about standard deviation, you should think in terms of distances to the mean, and not differences from the mean; otherwise you might be distracted by negative signs that aren't at all important. That is, if you're looking at this set: {1, 2, 3, 4, 5}, if you calculate the difference ("mean - x") for each x in the set, some of these differences will be positive, some negative. Those negative signs aren't important, since standard deviation is only based on distances from each element to the average - you can feed the numbers 2, 1, 0, 1, 2 into the standard deviation calculation to get the correct answer. Mind you, I've yet to see a real GMAT question that requires you to calculate standard deviation.

Since as you say those negative signs are not important, that is why the concept of Least is problematic, which is what I pointed out to him. If we squared the terms, the negative signs vanish so not important at all. Anyway the discussion was good b/c it helped reinforced concepts which may be needed on some Future SD problems. I agree they wouldn't ask u to calculate SD but if these principals are known, one can save a lot of time over some simple problems that appear difficult.

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by sanju09 » Wed Apr 08, 2009 3:02 am
d^2=[(a1 - a)^2 + (a2 - a)^2 +....+ (an - a)^2]/n

When we added 6 and 6, the numerator remained unchanged but the denominator increased, so, the new deviation is less than d.

Answer is E
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