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There is a set of number, the mean is m, the standard

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There is a set of number, the mean is m, the standard [#permalink] New post 13 May 2008, 05:04
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There is a set of number, the mean is m, the standard deviation is R, if add a number x to this set, is x<=R?
1) x=m
2) m<x<m+R

Dont know the answer, please explain.
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Re: DS- Standard Deviation [#permalink] New post 13 May 2008, 12:41
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I guess E

Consider two possible sets:

first set: (-101,-100,-99): mean (m) is -100, standard deviation is about 1
second set: (99,100,101): mean (m) is 100, standard deviation is about 1

1. or 2.
for first set x will be -100 or between -100 and -99 that are lesser than 1
for second set x will be 100 or between 100 and 101 that are greater than 1

1&2 are incompatible.
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Re: DS- Standard Deviation [#permalink] New post 13 May 2008, 14:24
am i wrong here or do the statement contradict each other?

1)x=m
2)x<m
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Re: DS- Standard Deviation [#permalink] New post 13 May 2008, 16:00
this one doesn't make sense to me.
1 and 2 cannot have opposite assumptions!
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Re: DS- Standard Deviation [#permalink] New post 13 May 2008, 18:00
I presume statement A is true.

When a no. which is equal to mean is added to the set, the std dev. does not change.

I am lost with the second statement.

please correct me if I am wrong.
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Re: DS- Standard Deviation [#permalink] New post 14 May 2008, 05:09
Value wrote:
There is a set of number, the mean is m, the standard deviation is R, if add a number x to this set, is r<=R? ( r=std dev of new set)
1) x=m
2) m<x<m+R

Dont know the answer, please explain.


Sorry Friends its r<=R not x<=R.
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Re: DS- Standard Deviation [#permalink] New post 14 May 2008, 05:22
Value wrote:
Value wrote:
There is a set of number, the mean is m, the standard deviation is R, if add a number x to this set, is r<=R? ( r=std dev of new set)
1) x=m
2) m<x<m+R

Dont know the answer, please explain.


Sorry Friends its r<=R not x<=R.



Maybe you could edit the original post also.
I did the whole thing before I read your correction.

Thanks
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Re: DS- Standard Deviation [#permalink] New post 14 May 2008, 05:28
With r<=R, I think its D.

Stt1: x=m, since x is exactly the mean it will leave the std. deviation unchanged. r=R. So the info is SUFF to answer.
Stt2: m<x<m+R => that x is larger than the mean. That means it is away from the center of the curve. So std. deviation must increase. (look at it like ( mean(x) - x(i) ) ^2 will add to the calcualtion of r). so r>R. So the info is SUFF to answer.
Thus, IMO: D
Re: DS- Standard Deviation   [#permalink] 14 May 2008, 05:28
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