Suppose the diameter at breast height in.) of trees of acertain type is normally distributed with mean 88 and sigma - 2.8. as suggested in the article "Simulating allarvester Forwarder Softwood Thinning" (Forest Products May 1997 36-41)
a. What is the probability that the diameter of a randomly selected tree will be at Wille least 10 in Will exceed 10 in.? h. What is the probability that the diameter of a randomly selected tree will exceed
in.? c. What is the probability that the diameter of a randomly selected tree will be
between
5 and 10 in?
20
d. What value c is such that the interval includes 98% of all diameter values? e. If four trees are independently selected, what is the probability that at least one has a diameter exceeding to in?
Answers
Given : diameter at breast height in.) of trees of ascertain type is normally distributed with mean 8.8 and sigma - 2.8.
To Find : probability that the diameter of a randomly selected tree will exceed at least 10 inches
probability that the diameter of a randomly selected tree will be between
5 and 10 in
Solution:
mean = 8.8
SD = 2.8
z score = ( value - Mean ) /SD
value = 10
=> z score = ( 10 - 8.8)/2.8 = 0.42857
from z score table : 0.6658
probability that the diameter of a randomly selected tree will exceed at least 10 inches = 1 - 0.6658 = 0.3342
between 5 and 10 in
value = 10
=> z score = ( 10 - 8.8)/2.8 = 0.42857
from z score table : 0.6658
value = 5
=> z score = (5 - 8.8)/2.8 = -1.357
from z score table : 0.0875
between 5 and 10 in = 0.6658 - 0.0875 = 0.5783
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