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

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p

x=0

step-by-step explanation: