Help with a normal distribution problem

Xanius

New Member
I'm studying for a test later today and I've run across a problem that I have no idea how to do.

The info I have is Mu = $6,000 sigma =$1200.

The problem is suppose we randomly sample 64 loans. What is the probability that at most 41 of these will have a dollar amount of \$6,000 or more?

I've tried to use the simple Z formula but that leaves me with 0/1200.
I have no idea how to do this, I can't use the other Z formulas because I don't have the information for them, and if I use mu for the sample mean, not sure if I can even though x-bar can be used as an unbiased estimator for mu, I still get a 0 in the numerator.

I know the answer, but not how to get to it.
The answer is .9878.

davers5

New Member
This is really a binomial probability question disguised as a normal distribution question. If you have a probability distribution that is normal with mean=6,000. Then the probability of a single observation being >=6,000 is .5. So what's the probability that there are at most 41 out of 65 observations are greater than 6,000? If you are familiar with binomial probabilities, you should be able to solve this easily.

Dave

Xanius

New Member
Ah, so the 0/1200 is the right answer for that and then I use that for the p in the binomial...makes sense once you know what it is.