Mean (μ) = 11.5 feet
Standard deviation (σ) = 1.7 feet
First we need to find the z-score for less than 13.5 feet.
The formula of z-score is : z = (X - μ)/σ
Here X= 13.5, so z = [tex] \frac{(13.5-11.5)}{1.7} [/tex]
z = [tex] \frac{2}{1.7} [/tex] = 1.18
P(X < 13.5) = P(z< 1.18) =P(z< (1.1 + .08)) = 0.8810 (from z-score table)
P(X< 13.5) = 88.1% (for making percentage from decimal, we need to multiply by 100)
So, the probability that a randomly selected tree is less than 13.5 feet tall = 88.1%
