My brain is oozing out of my head!

[ot180]

I am very lost. I have read my notes and text book but this teacher has me very confused. Can anyone fill me in?

The number of semester hours of college credit taken by candidates for a CPA exam has a distribution with a mean of 141 hours and a pop. standard deviation of 18 hours. Concider a random sample of 100 candidates.

I understand that if a pop has the N distribution then the sample mean of x bar of the n independent observations has the N distribution. However the next 2 are what i am completely stuck on. Any insight is VERY appreciated.

A)Find the probability that the sample mean credit hours for a sample of the 100 candidates exceeds 142.

B)Find the probabity that the sample mean credit hours for a sample of the 100 candidates lies between 90 and 120.

[JohnM]

Follow the same logic that is explained in our thread "The Vaunted Normal Distribution" in the Examples section, but z is computed as:

z = (x - mu) / (s/sqrt(n))

mu = 141
s = 18
n = 100

problem A:
x = 142
use the z table (normal distribution table) to find the probability of getting a z score above:
z = (142-141)/(18/sqrt(100))

problem B:
x = 90 and 120
use the z table (normal distribution table) to find the probability of getting a z score in between these two:
z = (90-141)/(18/sqrt(100))
z = (120-141)/(18/sqrt(100))