A company is preparing to launch a new web site to sell its products. From past

Question

A company is preparing to launch a new web site to sell its products. From past experience, they know that purchases at the site will occur as a Poisson process with on average 30 purchases every 10 minutes. What is the expected time in minutes before the first purchase after the site opens? Give your answer to four decimal places. What is the probability that more than 0.14 minutes pass before the first purchase? Give your answer to four decimal places. What is the expected time in minutes until the fifth purchase? Give your answer to four decimal places. What is the probability that the eighth purchase will occur between 1.33 minutes and 2 minutes after the site opens? Give your answer to four decimal places.

Explanation / Answer

In a homogeneous Poisson process, the probability that k events occur in an interval [t,t+t0] is:
P(N(t+t0)-N(t)=k) = e^(-lambda*t0)(lambda*t0)^k/k!

where lambda is the average number of events per time unit.
The time unit in your problem is one minute, and we know we have an average of 21 purchases every 10 minutes.
Lambda = 30/10 = 3 purchases/minute
----------------------------------------------------------------------------------------------------------------------------------------------------------

1. Expected time before the first purchase:
Probability that the first purchase (N1) occurs after t minutes is:
P(N1>t) = P(N(t)-N(0)=0) (because you must have exactly zero purchases in the interval [0, t]
P(N(t)-N(0)=0) = e^(-lambda*t)(lambda*t)^0/0! = e^(-lambda*t)
This is an exponential distribution with parameter lambda, and its expectation is 1/lambda.

You answer is t = 1/3 = 0.3333 minutes before the first purchase on average.

-----------------------------------------------------------------------------------------------------------------------------------------------

2. P(t>0.14) = P(N(0.14)-N(0)=0) = e^(-3*0.14) = 0.6570

------------------------------------------------------------------------------------------------------------------------------------------------

3. because of the memorylessness of the Poisson process, the inter-arrival time between two events are independent variables (with an exponential distribution and a mean of 1/lambda)
This means that you have a mean of 5/lambda minutes before the fifth arrival.
E(minutes before fourth purchase) = 5/3 = 1.6667 minutes

--------------------------------------------------------------------------------------------------------------------------------------------------------

4. P(1.33 <t(5th purchase)<2) = P(N(2)5) - P(N(1.33)4)
You must have at most three purchases at 1.33 minutes and at least 8 at 2 minutes.
P(N(2)5) =1- P(N(2)-N(0)=0) - P(N(2)-N(0)=1) - ....................- P(N(2)-N(0)=4)

P(N(2)5) = 1 - e^(-3*2)(3*2)^0/0! - e^(-3*2)(3*2)^1/1! - e^(-3*2) (3*2)^2 / 2! + ...................+ e^(-3*2) (3*2)^4 / 4!
P(N(2)5) = 1 - 0.0025 - 0.0149 - 0.0446 - 0.0892 - 0.1339 = 0.7149

P(N(1.33)4) = P(N(1.33)-N(0)=0) + P(N(1.33)-N(0)=1) +...............+ P(N(1.33)-N(0)=4)

P(N(1.33)4) = e^(-3*1.33)(3*1.33)^0/0! + e^(-3*1.33)(3*1.33)^1/1! + .............+ e^(-3*1.33)(3*1.33)^4/4!

P(N(1.33)4) = 0.0185 + 0.0738 + 0.1473 + 0.1959 + 0.1954 = 0.6308

Finally, P(1.33 <t(5th purchase)<2) = 0.7149 - 0.6308 = 0.0842

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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