Customers arrive at a suburban ticket outlet at the rate of 2 per hour on monday mornings. this can be described by a poisson distribution. selling the tickets and providing general information takes an average of 20 minutes per customer, and varies exponentially. there is 1 ticket agent on duty on mondays. what is the probability that an arriving customer will not have to wait for service

0.513246

Explanation:

Given the following;

Customer arrival rate = 2 per hour

selling the tickets and providing general information takes an average of 20 minutes per customer.

1 ticket agent

Since it is related by the poison distribution ;

λ = 2/hour

μ = 20 minutes per customer

μ = (60 ÷ 20) minutes

μ = 3 per hour

Average waiting time is calculated by;

λ ÷ [μ(μ - λ)]

Wait time = 2 ÷ [3(3-2)]

Wait time = 2 ÷ 3

= 0.667hours

Therefore, probability that customer will not have to wait for service, that is wait time, x = 0

P(x; μ) = (e^-μ) (μ^x) / x!

P(0; 0.667) = (0.513246) (1) / 0!

P(0; 0.667) = 0.513246 / 1

P(0; 0.667) = 0.513246