Assume that the diameter at breast height (in.) of trees of a certain type is no

Question

Assume that the diameter at breast height (in.) of trees of a certain type is normally distributed with a mean of 8.8 in. and a standard deviation of 2.8 in.

a What is the probability that a randomly selected tree will be at least 12 in. in diameter?

b What is the probability that the diameter of a randomly selected tree will be between 5 and 10 in.?

c 15% of all tree diameters for this type of tree exceed what value?

d If four trees are independently selected, what is the probability that at least one has a diameter exceeding 12 in.?

Explanation / Answer

a)P(X>12)=P(Z>(12-8.8)/2.8)=P(Z>1.14)=0.1271

b)P(5<X<10)=P((5-8.8)/2.8<Z<(10-8.8)/2.8)=P(-1.36<Z<0.43)=0.6664-0.0869 =0.5795

c)

for top 15 percentile ; critical z =1.04

hence corresponding value =mean +z*std deviation =8.8+1.04*2.8=11.71

d)

P(at least one has dia, exceeding 12 in)=1-P(all have dia. less than 12)=1-(1-0.1271)4 =0.4194

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