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A certain characteristic in a large population has a

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A certain characteristic in a large population has a [#permalink] New post 05 Sep 2008, 20:03
A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is more than m-d?
A. 16%
B. 32%
C. 48%
D. 84%
E. 92%
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Re: normal distribution [#permalink] New post 05 Sep 2008, 20:24
arjtryarjtry wrote:
A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is more than m-d?
A. 16%
B. 32%
C. 48%
D. 84%
E. 92%


since out of 100 ,68 lie within m-d deviation,32 must lie outside or on the mean !!! A Vs B is answer
C,D,E get eliminated since we need values less than or equal to 32 .

since distribution is symmetric about mean m.16% will be the answer
IMO A
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Re: normal distribution [#permalink] New post 06 Sep 2008, 05:12
spriya wrote:
arjtryarjtry wrote:
A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is more than m-d?
A. 16%
B. 32%
C. 48%
D. 84%
E. 92%


since out of 100 ,68 lie within m-d deviation,32 must lie outside or on the mean !!! A Vs B is answer
C,D,E get eliminated since we need values less than or equal to 32 .

since distribution is symmetric about mean m.16% will be the answer
IMO A


IMO D is the answer.

spriya i think D cannot be eliminated by default.
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Re: normal distribution [#permalink] New post 06 Sep 2008, 08:48
I arrived at D as well because % of distribution above m-d is

( % of distro between m-d and m+d ) + ( % of distro between m+d and 100)

Symmetric mean?? Does life with GMAT has to get so complex??
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Re: normal distribution [#permalink] New post 06 Sep 2008, 09:05
vote for D
question asks for distribution more than m-d line..not less ...
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Re: normal distribution [#permalink] New post 06 Sep 2008, 09:26
slippery one
D is my answer, however on first thought i had jumped the gun to 16
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Re: normal distribution [#permalink] New post 06 Sep 2008, 19:10
i also got D. so by consensus...
we can vote D as the winner????
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Re: normal distribution [#permalink] New post 06 Sep 2008, 23:39
arjtryarjtry wrote:
i also got D. so by consensus...
we can vote D as the winner????

I believe the distribution outside deviation m-d will exclude those within m-d and with m-d hence 16%

anyways whats the OA?
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Re: normal distribution [#permalink] New post 07 Sep 2008, 00:08
Calling Walker, et al.

Could you explain this?
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Re: normal distribution [#permalink] New post 08 Sep 2008, 01:39
spriya wrote:
arjtryarjtry wrote:
i also got D. so by consensus...
we can vote D as the winner????

I believe the distribution outside deviation m-d will exclude those within m-d and with m-d hence 16%

anyways whats the OA?


The question asks for % of distribution more than m-d. More than m-d is 68% contained between m-d and m+d plus additional 16% between m+d and 100.
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Re: normal distribution [#permalink] New post 08 Sep 2008, 01:57
arjtryarjtry wrote:
A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is more than m-d?
A. 16%
B. 32%
C. 48%
D. 84%
E. 92%


I agree with the D answers.

The question asks what percent of the distribution is more than m-d. If you picture it like a typical bell curve, m-d is on the left and the question asks for everything to the right of that point.

The population within one standard deviation is 68% (both to the right and the left of the center of the bell curve.) That means half of 68% accounts for the distance from m-d to m: 34%. Then you need everything else that's greater than m, which is half of the entire distribution, or 50%. 50+34 = 84%.

[Not meaning to confuse, but in my head I actually something similar to icandy's way:
( % of distro between m-d and m+d ) + ( % of distro between m+d and 100)
which is same as
( % of distro between m-d and m+d ) + ( half of the % of distro outside of m+d and m-d)
which is (68%) + ((100-68)/2)% = 84%]
Re: normal distribution   [#permalink] 08 Sep 2008, 01:57
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