Suppose the diameter at breast height (in.) of trees of a certain type is normal

Question

Suppose the diameter at breast height (in.) of trees of a certain type is normally distributed with = 8.7 and = 2.5.

(a) What is the probability that the diameter of a randomly selected tree will be at least 10 in.? Will exceed 10 in.?
At least 10 = ........ (round to 4 decimals)


(c) What is the probability that the diameter of a randomly selected tree will be between 5 and 10 in.?
(round to 4 decimals)
(d) What value c is such that the interval (8.7 - c, 8.7 + c) 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 10 in.?
(round to 4 decimals)

Explanation / Answer

a) At least 10 = =P(X>10)=1-P(X<10)=1-P(Z<(10-8.7)/2.5)=1-P(Z<.52)=1-0.6985 =0.3015

c)probability that the diameter of a randomly selected tree will be between 5 and 10 in=P(5<X<10)

=P(-1.48<Z<0.52)=0.6985-0.0694 =0.6290

d) for 98% CI ; z =2.3263

hence c =z*std error =5.8159

e)

for sample size n=4 ; std error =std deviation/(n)1/2 =2.5/(4)1/2 =1.25

At least 10 = =P(X>10)=1-P(X<10)=1-P(Z<(10-8.7)/1.25)=1-P(Z<1.04)=1-0.8508 =0.1492

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.