3. An art collector travels to art auctions once a month on the average. Each tr

Question

3. An art collector travels to art auctions once a month on the average. Each trip is guaranteed to produce one purchase. The time between trips is exponentially distributed. Determine a. The probability that no purchase is made in a 3-month period b. The probability that no more than 8 purchases are made per year. c. The probability that the time between successive trips will exceed 1 month. 4. The time between arrivals at E&L; restaurant is exponential with mean 5 minutes. The restaurant opens for business at 11:00 A.M. Determine: a. The probability of having 10 arrivals in the restaurant by 11:b A.M. given that 8 customers arrived by 1 1:(1-8) A.M. b. If the last customer arrived at 11.25 A.M, find the probability that a new customers will arrive between 11:28 and 11:33 A.M.

Explanation / Answer

=EXPONDIST(25,B29,TRUE)-EXPONDIST(5,B29,TRUE)

Ans 1 a mean 1 1/mean =1/1 P(X<=3) =EXPONDIST(3,B2,TRUE) P(1<=X<=3) =EXPONDIST(3,1,TRUE)-EXPONDIST(1,1,TRUE) Ans 1 b mean 1 1/mean 1 P(X<=8) =EXPONDIST(8,B9,TRUE) P(1<=X<=8) =EXPONDIST(8,1,TRUE)-EXPONDIST(1,1,TRUE) Ans 1 c mean 1 1/mean 1 P(X<=1) =EXPONDIST(1,B16,TRUE) P(0<=X<=1) =EXPONDIST(1,1,TRUE)-EXPONDIST(0,1,TRUE) Ans 2 a mean 5 1/mean =1/5 P(X<=10) =EXPONDIST(10,B22,TRUE) P(8<=X<=10) =EXPONDIST(10,5,TRUE)-EXPONDIST(8,5,TRUE) Ans 2 b mean 5 1/mean =1/5 P(X<=25) =EXPONDIST(25,B29,TRUE) P(5<=X<=25)

=EXPONDIST(25,B29,TRUE)-EXPONDIST(5,B29,TRUE)

Ans 1 a mean 1 1/mean 1 P(X<=3) 0.950212932 P(1<=X<=3) 0.318092373 Ans 1 b mean 1 1/mean 1 P(X<=8) 0.999664537 P(1<=X<=8) 0.367543979 Ans 1 c mean 1 1/mean 1 P(X<=1) 0.632120559 P(0<=X<=1) 0.6 Ans 2 a mean 5 1/mean 0.2 P(X<=10) 1 P(8<=X<=10) 0.0 Ans 2 b mean 5 1/mean 0.2 P(X<=25) 1.0 P(5<=X<=25) 0

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

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