Question

An internet service provider offers special discounts to every third connecting customer. its customers connect to the internet according to a poisson process with the rate of 5 customers per minute. compute:(a) the probability that no offer is made during the first 2 minutes (b) expectation and variance of the time of the first offer

a) 0.0025, Expectation: 0.2 minutes, Variance: 0.04 minutes
b) 0.0045, Expectation: 0.3 minutes, Variance: 0.05 minutes
c) 0.0067, Expectation: 0.5 minutes, Variance: 0.25 minutes
d) 0.0089, Expectation: 0.8 minutes, Variance: 0.16 minutes

Answer 1

The probability that no offer is made during the first 2 minutes is e^(-10). The expectation and variance of the time of the first offer are 0.2 minutes and 0.04 minutes, respectively.

Step-by-step explanation:

The probability that no offer is made during the first 2 minutes can be calculated using the Poisson distribution. The rate parameter of the Poisson distribution is 5 customers per minute, so the average number of customers in a 2 minute interval is 10. The probability that no offers are made in this interval can be calculated using the formula P(X=0) = e^(-10) * (10^0 / 0!) = e^(-10).

The expectation of the time of the first offer can be calculated as the reciprocal of the rate parameter, which is 1/5 = 0.2 minutes. The variance of the time of the first offer can be calculated as the reciprocal of the rate parameter squared, which is 1/(5^2) = 0.04 minutes. Therefore, the correct answer is (a) 0.0025, Expectation: 0.2 minutes, Variance: 0.04 minutes.