... time’=0 .90 9 090 9 091 6 6‘Mean service duration’=0 .90 9 090 9 091 , ‘Std dev of service duration’=0 .90 9 090 9 091 6 ‘mean cycle length’=11.4485 597 9 6 ‘estimate of mean line length’=10.32733 293 95 % Confidence interval ... 6 6 6 Appendices ‘Mean interarrival time’=1.,‘Std dev of interarrival time’=0 .90 9 090 9 091 6‘Mean service duration’=0 .90 9 090 9 091 , Std dev of service duration’=0 .90 9 090 9 091 6 ‘mean cycle length’=10.86151154 ... 2.5,4.4,2.5,4.8,3.5,infinity,infinity,infinity, 6 infinity,infinity]: 6 print(" _INPUT 6 DATA "); 6 seeda:=87341:seedb:=64287: 6 print("Arrival and service seeds"=seeda,seedb); 6 print("Run...