Normal Probability

Question

Let  represent the full height of a certain species of tree. Assume that  has a normal probability distribution with a mean of 87.4 ft and a standard deviation of 4.8 ft.A tree of this type grows in my backyard, and it stands 76.8 feet tall.

1-  Find the probability that the height of a randomly selected tree is as tall as mine or shorter. Enter your answer as a decimal rounded to 4 decimal places if necessary. = My neighbor also has a tree of this type growing in her backyard, but hers stands 77.3 feet tall.

2- Find the probability that the full height of a randomly selected tree is at least as tall as hers. Enter your answer as a decimal rounded to 4 decimal places if necessary. = 

Answer 1

P(X<76.8) = P(Z<((76.8-87.4)/4.8) = P(Z<-2.2083) = Normalcdf(-10,000, -2.2083, 0, 1) = 0.0136

or Normalcdf(-10,000, 76.8, 87.4, 4.8) = 0.0136

P(X>77.3) =P(Z>(77.3-87.4)/4.8) = P(Z>-2.104) = Normalcdf(-2.104, 10,000, 0,1) = 0.9823

or Normalcdf(77.3, 10,000, 87.4, 4.8) = 0.9823