Question 1121626
z=(x-mean)/sd
for a, z>(70-74)/12.5 or z>-0.32.  That probability is 0.6255

b.  z=(x-mean)/s/sqrt(n) or z>-4*sqrt(25)/12.5 or -1.6  That probability is 0.9452

c.  If we can assume the population is normally distributed, then any sample size is appropriate.  Thirty is not magic.  If the population is significantly skewed, one may need a lot more than 30 or even a nonparametric test.  If the population is normally distributed, 10 would be fine.