How to solve this kind of standard deviation sums???

[raunekk]

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 less than m + d ?
A. 16%
B. 32%
C. 48%
D. 84%

E. 92%

[Retake_GMAT]

Since it is symmetric distribution, i assume it is normal distribution.

100 - ((34 +16 +2) - (34))=84%

Hence it is 84%. ans D

[sanjeev_sharma]

Since this is a normal distribution of mean, the % of population <= 68 + ((100-68)/2) = 68 + 32/2 = 68 + 16 = 84%

[GmatG]

Originally Posted by sanjeev_sharma

Since this is a normal distribution of mean, the % of population <= 68 + ((100-68)/2) = 68 + 32/2 = 68 + 16 = 84%

Agree to Sanjeev Sharma IMO 84%

[raunekk]

Originally Posted by Retake_GMAT

Since it is symmetric distribution, i assume it is normal distribution.

100 - ((34 +16 +2) - (34))=84%

Hence it is 84%. ans D

Can you pls explain as to how to start for this...

There is a method in priceton review to solve such kinda problems..

but i am not able to apply it over here..

[Anson800]

IMO D. Search on forum. Somwhere you will get graph of Standard deviation.
Basically, values are distributed in 2-14-34-0-34-14-2 way. 0 corr to mean 34 corr to 1 std dev and 14 to 2 std dev