Question having to do with normal distributions

Hi, I'm trying to figure out this problem, but I'm having a hard time.

Two stores sells landscaping rocks. At the first store rocks weight an average of 22 pounds, with a standard deviation of 2.5 pounds. At the second store the rocks are smaller, with a mean of 18 pounds and SD of 2 pounds. You select a rock at random from each store.

a. What are the mean and standard deviation of the difference in weights of the melons.
Mean of difference = 22-18=4
Standard deviation of difference = sqrt(2.5^2+2^2) = 3.202
Is this right?

b. Assume that the difference in weights has a normal distribution. What is the probability that the rock you get at the first store is heavier?

Any help with part b? Thanks.
Okay, so I think I figured out part B. Can someone confirm my answer?

Let M = weight of rock from store 1
Let W = weight of rock from store 2
Want prob that M>W
D = difference in weight between M and W
Want probability that M-W is greater than 0
Expected value of difference D is 22-18 = 4
From part A the standard deviation is 3.20
We want to know the probability that the difference is greater than 0 pounds, which has a z-score of -1.25 using the SD from part A. Since the difference in weight is distributed evenly, the probability that the difference is greater than 0 is .8944, or 89.44%.

Is this right?