Posted by Robbie on .
A normal distribution has a mean of 36 and standard deviation of 4. What proportion of the
distribution consisted of scores between x=34 and x=38?
a, 0.6915
b, 2(0.3085)=0.6170
c, 0.1915
d, 2(0.1915)=0.3930

statistics 
PsyDAG,
Z = (scoremean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.