You must to google a bit.... a found this example on the net... see more details here:

http://www.intmath.com/Counting-probability/14_Normal-probability-distribution.php
It was found that the mean length of 100 parts produced by a lathe was 20.05 mm with a standard deviation of 0.02 mm. Find the probability that a part selected at random would have a length

(a) between 20.03 mm and 20.08 mm

(b) between 20.06 mm and 20.07 mm

(c) less than 20.01 mm

(d) greater than 20.09 mm.

Answer
*X* = length of part

(a) 20.03 is 1 standard deviation below the mean;

20.08 is

standard deviations above the mean

So the probability is 0.7745.

(b) 20.06 is 0.5 standard deviations above the mean;

20.07 is 1 standard deviation above the mean

So the probability is 0.1498.

(c) 20.01 is 2 s.d. below the mean.

So the probability is 0.0228.

(d) 20.09 is 2 s.d. above the mean, so the answer will be the same as (c),

*P*(*X* > 20.09) = 0.0228.