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by gmatnmein2010 » Mon Feb 15, 2010 8:42 pm
A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68% of the distribution lies one standard deviation d of the mean, what percent of the distribution is less than m+d?
(A) 16%
(B) 32%
(C) 48%
(D) 84%
(E) 92%
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by komal » Mon Feb 15, 2010 9:04 pm
gmatnmein2010 wrote:A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68% of the distribution lies one standard deviation d of the mean, what percent of the distribution is less than m+d?
(A) 16%
(B) 32%
(C) 48%
(D) 84%
(E) 92%
Since it is symmetric about m, it means, data less than m is 50% and data greater than m is also 50%.
For distribution betwen one std dev d, the distribution from m+d to m-d would also be symmetric about m and thus would contain two equal, symmetrical halves.
Thus, each of the area from m to m+d and to m-d would be 34% or 68/2=34%.

Thus, for data to be less than m+d, we have

Data from m+d to m = 34%
Data less than m = 50%
Thus, total 84%

(D) is correct
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by shashank.ism » Mon Feb 15, 2010 9:04 pm
gmatnmein2010 wrote:A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68% of the distribution lies one standard deviation d of the mean, what percent of the distribution is less than m+d?
(A) 16%
(B) 32%
(C) 48%
(D) 84%
(E) 92%
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