Statistics: How to solve this Normal Sampling Distribution Question

A manufacturing process is designed to produce bolts with a 0.75-in. diameter. Once each day, a random sample of 36 bolts is selected and the diameters recorded. If the resulting sample mean is less than 0.735 in. or greater than 0.765 in., the process is shut down for adjustment. The standard deviation for diameter is 0.03 in. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an x in the shutdown range when the true process mean really is 0.75 in. Round all your intermediate calculations to four decimal places. Round the answer to four decimal places.)

I changed the numbers so that I can understand from your explanation how to solve my actual homework problem.

Thank you

Hmm...these numbers you gave us are awfully convenient and really simplify the calculations. Are you sure you made them up? =)

Let Xbar = sample mean of bolt diameter.
Then since n = 36 > 30, by the Central Limit Theorem, we have:
Xbar ~ Normal(μ, 0.03^2/36).

Since the manufacturing line will be shut down unnecessarily, assume that μ = 0.75.
Then we want:
P(Xbar < 0.735) + P(Xbar > 0.765)
= P(Z < (0.735 - 0.75)/(0.03/sqrt(36))) + P(Z > (0.765 - 0.75)/(0.03/sqrt(36)))
= P(Z < -3) + P(Z > 3)
= 0.00135 + 0.00135
= 0.0027