gre standard deviation probability 
I have solved it ,but my answer is wrong 
above mean
mean + n S.d 
Below mean 
mean-n s.d
S.d' standard deviation and 'n' is number of time s.d 
 A: Answer
Qty B is greated because 


*

*the lengths of the interval is the segments are the same, 

*the distance between the mean $3.05$ and segment B is smaller


Deviation is irrelelvant as long as it is different from $0$.
You do not need to do any computation to prove it. I did not have to right anything on paper or use the computer or calculator to give this answer.
Proof
As you probably know, the density of the normal variable is $\rho(x) := \frac{1}{\sqrt{2\pi}\sigma} \exp{-\frac{(x-m)^2}{2 \sigma^2}}$
Then $$Pr(X \in [4,5]) - Pr(X \in [1,2])= \int_4^5 \rho(x)dx - \int_1^2 \rho(x)dx \\= \frac{1}{\sqrt{2\pi}\sigma}\left(\int_{4-3.05}^{5-3.05} \exp{-\frac{x^2}{2 \sigma^2}}dx - \int_{1-3.05}^{2-3.05} \exp{-\frac{x^2}{2 \sigma^2}}dx\right)\\= \frac{1}{\sqrt{2\pi}\sigma}\left(\int_{0.95}^{1.95} \exp{-\frac{x^2}{2 \sigma^2}}dx - \int_{1.05}^{2.05} \exp{-\frac{x^2}{2 \sigma^2}}dx\right)\\= \frac{1}{\sqrt{2\pi}\sigma}\left(\int_{0.95}^{1.05} \exp{-\frac{x^2}{2 \sigma^2}}dx - \int_{1.95}^{2.05} \exp{-\frac{x^2}{2 \sigma^2}}dx\right)\\$$
The last expression is obviously positive.
A: The lengths of the intervals are the same ($1$). In case A, the bound closer to the mean ($2$) is $1.05$ away, while in case B, the bound closer to the mean ($4$) is just $0.95$ away. If you sketch the normal distribution curve, it should be fairly obvious that equal intervals at the tails give rise to higher areas under the curve the closer the bounds are to the mean. Hence Quantity B is the larger.
