The st. joe company grows pine trees and the average annual increase in tree diameter is 3.1 inches with a standard deviation of 0.5 inch. a random sample of n = 50 trees is collected. what is the probability of the sample mean being less the 2.9 inches?

Solution: We are given:

μ=3.1,σ=0.5, n=50

We have to find P (Mean <2.9)

We need to first find the z score

z = (xbar-μ) / (σ/sqrt (n))

= (2.9-3.1) / (0.5/sqrt (50))

= (-0.2) / 0.0707

=-2.83

Now we have to find P (z<-2.83)

Using the standard normal table, we have:

P (z<-2.83) = 0.0023

Therefore the probability of the sample mean being less the 2.9 inches is 0.0023