Hey yvonnehr.
If this is a waiting time that is continuous then you have might want to check the exponential distribution:
Exponential distribution - Wikipedia, the free encyclopedia
I don't know what formula to use to solve this problem. It sounds like a Geometric Distribution but I'm not sure.
After repeated observations, it has been determined that the waiting time at the drive-through window of a local bank is skewed left, with a mean of 3.5 minutes and a standard deviation of 1.9 minutes. A random sample of 100 customers is to be taken. What is the probability that the mean of the sample will exceed 4 minutes?
The answer the teacher provided is 0.0042.
Thank you.
Hey yvonnehr.
If this is a waiting time that is continuous then you have might want to check the exponential distribution:
Exponential distribution - Wikipedia, the free encyclopedia
By the Central Limit Theorem, the mean is approximately Normal with mean 3.5 and a standard deviation of . (That the waiting time is skewed left is irrelevant.)
So what is the probability that a Normal random variable with mean 3.5 and standard deviation (as above) is greater than 4?