Suppose that customers arrive at a check-out according to a Poisson process with

Question

Suppose that customers arrive at a check-out according to a Poisson process with mean lambda = 12 per hour. What is the probability that we will have to wait longer than 15 minutes to see the first customer? Suppose we have already waited for 15 minutes, still no customers arrived. Conditioning on this, what is the probability that we will have to wait longer than 15 minutes to see the first customer? Denoting Y as the time between the 10th and the 15th customer. What is the distribution of y? Find its mean and variance.

Explanation / Answer

a) average time for a customer = 12/60 = 0.2

probability wait longer than 15 minutes = P(X>15) =exp(-0.2*15)=0.05

b)

average time per customer

P(X>0|X>15) = P(X>15) = 0.05

This is memory less property

c) time between 10 th 15 th = exp(-0.2*10)-(exp(-0.2*15) = 0.086

Mean = 0.2

variance = 0.04

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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