Answer to Question #86116 in Statistics and Probability for Anand

Question

In a railway yard goods trains arrive that at the rate of 30 trains per day. Assuming
that the inter-arrival time follows an exponential distribution and the service time
distribution is also exponential with an average of 36 minutes, calculate the
following :
(i) The average number of trains in the queue.
(ii) The probability that the queue size is greater than or equal to 10

Answer 1

(i) Average number or trains in the queue is

E(m)=(λ2 )/(μ(μ-λ))

λ=30/(60x24)=1/48 trains per minute

E(m)=((1/48)² )/(1/36(1/36-1/48))=108/48=2.25 or rounghly 2 trains

(ii) The probability that number or trains in it system exceeds 10

P(≥10)=P¹⁰ =(λ/μ)¹⁰=((1/48)/(1/36))¹⁰=(0.75)¹⁰ =0.06