Erlang Distribution Parameters and Carwash Example
Explore the Erlang distribution with parameters k=2 and λ=3, and learn about a carwash example involving outside washing, inside cleaning, and waiting time analysis. Understand how to calculate average, variance, and probability of waiting times. Integration methods to solve related problems are discussed.
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= = = 0.5 = = 0.5 3 0.5 1 (1 3) | ( ) 0.38 3 1 P X e e e t t t 1 20 = = = ( ) 30 0.38 30 P X e dt t 3 1 60 1 20 60 1 60 (2 30) (2 30) 60! 60 e = = = ( 60) P N 2
Gamma randomVariable = 1 x ( ) ( ) x e = ( ) 1 x where x e dx ( ) f x ( 0) x 0 = = 1 n x ( ) n ( 1)! x e dx n When is an integer number say n it can shown that 0
1 k R k = = = ( ) T Erlang E T k Carwash Example A carwash consists of two sections: outside washing and inside cleaning. The average service time at each of these parts is 20 minutes. What is the average and variance of waiting time for a car which start to get its service? What is the probability that the waiting time for this car be between 30 and 45 minutes? We knew that when each part of total service point governs by exponential with mean 20 minutes or hours (parameter =3), the total service time will govern by an Erlang distribution with parameters k=2 and = 3 . Then 1 = 1 3
Erlang With parameters k=2 and =3 Outside Inside Exponential With parameter =3 Exponential With parameter =3 k k = = = = [ ] E T 40 , 10.33 , hours or Minutes Var hours or Minutes 2 3 2 9 2 2 1 1 3 k T T ( ) 3 (3 ) (2 1)! T k e T e = = = 3 T ( ) f T ( ) 9 Since Then f T Te ( 1)! 3 4 = = 3 T (30 45) ( ) 9 P T P T Te dT 3 4 1 2 1 2 Integration by part can help us to solve the above integration. Then lets assume that = = = = 3 3 T T = , u T and dv e dT then du dT and v e 1 ( ) udv uv vdu 3 3 4 3 4 3 4 1 2 3 4 1 2 3 4 1 2 = = 9( )( ) 3 3 3 3 3 T T T T T 9 9[ | ] 3 | | Te dT Te e dT Te e 1 1 1 1 3 3 3 3 1 2 1 2 3 4 3 4 9 4 3 2 1 2 1 2 3 3 3 3 = + + = + = 3 3 0.215 e e e e e e 3 4 13 4 5 2 1 2
