Assume that a population is normally distributed with a mean of 100 and a standard deviation of 15.
2 answers
What is your question?
Yes, it would be unusal for the mean of a sample of 3 to be 115 or more because there is only a 4.18% chance for this to happen.
P(X >= 115)
= P(X >= (115 - 100) / sqrt(15^2 / 3))
= P(X >= 1.732050808)
= 0.0418
P(X >= 115)
= P(X >= (115 - 100) / sqrt(15^2 / 3))
= P(X >= 1.732050808)
= 0.0418