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Re: There is a set of number, the mean is m, the standard [#permalink]
this one doesn't make sense to me.
1 and 2 cannot have opposite assumptions!
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Re: There is a set of number, the mean is m, the standard [#permalink]
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: There is a set of number, the mean is m, the standard [#permalink]
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: There is a set of number, the mean is m, the standard [#permalink]
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: There is a set of number, the mean is m, the standard [#permalink]
With 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



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