Let XX represent the full height of a certain species of tree.
Assume that XX has a normal probability distribution with a mean of
107 ft and a standard deviation of 5 ft.
A tree of this type grows in my backyard, and it stands 91.5 feet
tall. Find the probability that the height of a randomly selected
tree is as tall as mine or shorter.
P(X<91.5)P(X<91.5) =
My neighbor also has a tree of this type growing in her backyard,
but hers stands 92.5 feet tall. Find the probability that the full
height of a randomly selected tree is at least as tall as
hers.
P(X>92.5)P(X>92.5) =
Solution :
Given that ,
mean = = 107
standard deviation = =5
(a)P(x <91.5 ) = P[(x - ) / < (91.5 - 107) / 5]
= P(z <-3.1 )
Using z table,
answer =0.0010
(b)P(x >92.5 ) = 1 - p( x<92.5 )
=1- p P[(x - ) / < (92.5 - 107) /5 ]
=1- P(z < -2.9)
Using z table,
= 1 -0.0019
answer=0.9981
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