Answer to Question #106151 in Statistics and Probability for simran

Question

cassette player repairman finds that the time spent on his job has an exponential distribution
with mean 15 min. If he repairs sets in the order in which they came in, and if the arrival of sets
is approximately Poisson with an average rate of 18 per 9 hours a day, what is repairman’s
expected idle time each day? How many jobs are ahead of the set just brought in?

Answer 1

From the data of the problem, we have

$\lambda=18/9=2\ \text{sets per hour,}$

$\mu=1(60/15)=4\ \ \text{sets per hour}$

(a) Expected idle time of repairman each day = Number of hours for which the repairman remains busy in an 9-hour day (traffic intensity) is given by    

$9(\lambda/\mu)=9(2/4)=4.5\ hours$

Hence, the idle time for a repairman in an 9-hour day will be:

$9-4.5=4.5\ hours$

(b) Expected (or average) number of cassette sets in the system

$L_s={\lambda \over \mu-\lambda}={2\over 4-2}=1\text{ cassette set}$