Question

Let X represent the full height of a certain species of tree. Assume that X has a normal probability distribution with a mean of 228.3 ft and a standard deviation of 7.9 ft. A tree of this type grows in my backyard, and it stands 207 feet tall. Find the probability that the height of a randomly selected tree is as tall as mine or shorter. P ( X < 207 ) = My neighbor also has a tree of this type growing in her backyard, but hers stands 212.5 feet tall. Find the probability that the full height of a randomly selected tree is at least as tall as hers. P ( X > 212.5 ) = Enter your answers as decimals accurate to 4 decimal places.

Answer 1

Solution:-

Mean = 228.3, S.D = 7.9

a) The probability that the height of a randomly selected tree is as tall as mine or shorter. P ( X < 207 ) = 0.0035.

x = 207

By applying normal distribution:-

z = - 2.696

P(z < - 2.696) = 0.0035

b) The probability that the full height of a randomly selected tree is at least as tall as hers. P ( X > 212.5 ) = 0.9772 .

x = 212.5

By applying normal distribution:-

z = - 2.0

P(z > - 2.0) = 0.9772