Find the probability that the mean annual preciptiation will be between 32 and 34 inches. variable is normally distributed

Supposing, for the sake of illustration, that the mean is 31.2 and the std. dev. is 1.9.

This probability can be calculated by finding z-scores and their corresponding areas under the std. normal curve.  
                                                                     34 in - 31.2 in
The area under this curve to the left of z = = 1.47 (for 34 in)
                                                                           1.9
                                                                      32 in - 31.2 in
and that to the left of 32 in   is               z = = 0.421
                                                                             1.9

Know how to use a table of z-scores to find these two areas?  If not, let me know and I'll go over that with you.

My TI-83 calculator provided the following result:

normalcdf(32, 34, 31.2, 1.9) = 0.267  (answer to this sample problem)

There is no association.

Provide an equation