Finding probability using CDF for standard normal RV

Hey all, first time poster, but I'm sure this won't be my only visit.

I'm in a discrete mathematics class, and am really stuck on this one problem. Basically, it goes as follows. There is a place, Bob's metal shed, that produces long metal tubes of size L for their customers. The customer specifies L, and n, the number of pieces, and they will receive n pieces of size between L-.5cm and L+.5cm. The length of the pieces cut has a variance of 1/16, and the cut length of the pieces follows a normal distribution. Produced pieces outside of the L-.5 to L+.5 range are defectives and are thrown out. The total number of pieces produced, both defective and not, is m.

So... I'm asked to find the probability that the machine produced a piece which can be delivered to the customer, using the cumulative distribution function for the standard normal RV. I know that the sigma^2 parameter is the variance, 1/16, and that mu, the expected value, would be mp, where p is the probability of an acceptable piece being made. p would be equal to n/m, and therefore mu is m*(n/m)=n. This is about where I get stuck... How do I go from that, to using the CDF to find the probability of one piece being acceptable? Preemptive thanks for the help!

Re: Finding probability using CDF for standard normal RV

Welcome to the forum.

The CDF of any normal variable can be expressed through the standard normal CDF . See the Wiki page and also this similar thread.

Since in your example the maximum allowed deviation 0.5cm is twice the standard deviation , you need to find . If you express through , you'll get rid of and . See also the Wiki section on confidence intervals.

Even though this problem is from discrete mathematics class, the problem itself is purely probability theory. In the future, you'll have more chances to get a good reply if you post it to either basic or advanced statistics and probability section.