So this is what I came up with, please tell me if I'm wrong...
P(x>75)=
P(z>(75-70)/3)=
P(z>1.68)=
0.0465
The daily sale of lunch is a random variable. Let us call it X. X ~ N (70,9). What is the probability that in a given day the restaurant sells more than 75 lunches?
Is this telling me that the population average is 70 lunches with variance of 9?
Your method is OK but the details leave a bit to be desired.
5/3 = 1.67 (correct to two decimal places) not 1.68.
Also, I don't see the point in rounding z to two decimal places and then giving a probability to four decimal places.
Either use a more accurate value of z or round the probability to less than four decimal places. (Personally I'd do the former).