ENGR 323 Beautiful Homework #9 1 of 5 Johnston Problem b) The standard deviation of the time between successive arrivals
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1 ENGR 33 Bautiful Homwork #9 of 5 Johnston Problm 4.58 PROBLEM STATEMENT Lt th tim btwn two succssiv arrivals at th driv-up window of a local bank. If has an xponntial distribution with (which is idntical to a standard gamma distribution with α ), comput th following: a) Th xpctd tim btwn two succssiv arrivals b) Th standard dviation of th tim btwn succssiv arrivals c) 4) d) 5) Th givn problm did not provid th units of tim. Lt's assum th local bank is Humboldt Bank and th distribution rat () is arrival pr hour. EPONENTIAL DISTRIBUTION Th xponntial pdf is a spcial cas of th gnral gamma pdf in which α and β has bn rplacd by /. (s th gnral gamma pdf in th txt on pg 67) Figur is a graph of th xponntial pdf with th paramtr arrival pr hour. Exponntial pdf x x f ( x; ) whr > othrwis.8 f(x;) x tim btwn two succssiv arrivals (hours) Figur : Exponntial Probability Dnsity Function with
2 ENGR 33 Bautiful Homwork #9 of 5 Johnston Problm 4.58 Th xponntial cdf is found by intgrating th xponntial pdf. Figur is a graph of th xponntial cdf with th paramtr arrival pr hour. Exponntial cdf F( x; ) x x x F (x; ) x tim btwn two succssiv arrivals Figur : Exponntial Cumulativ Dnsity Function with Th man and varianc ar dtrmind by using th gnral gamma xprssions, whr α and β /. Man µ αβ Varianc αβ σ
3 ENGR 33 Bautiful Homwork #9 3 of 5 Johnston Problm 4.58 PART (a) SOLUTION tim btwn two succssiv arrivals (assumd units is hours) ~ Exp ( ) Th xpctd tim btwn two succssiv arrivals is: E ( x) µ hour Answr: µ hour PART (b) SOLUTION tim btwn two succssiv arrivals (assumd units is hours) ~ Exp ( ) Th varianc of th tim btwn succssiv arrivals is: σ hour Th standard dviation of th tim btwn succssiv arrivals is: σ σ Answr: σ hour hour PART (c) SOLUTION tim btwn two succssiv arrivals (assumd units is hours) ~ Exp ( ) What is th probability that th tim btwn two succssiv arrivals is lss than or qual to four hours? This probability can b dtrmind in svral diffrnt ways. On way is to intgrat th xponntial pdf: 4) 4 4 x () () x
4 ENGR 33 Bautiful Homwork #9 4 of 5 Johnston Problm 4.58 An asir mthod is to calculat th cdf: 4) F( x; ) F(4;) x ()(4).987 Answr: 4).987 Poisson Mthod If th waiting tim () is xponntially distributd, thn th numbr of arrivals is a Poisson distribution. (s Poisson distribution in txt on pg 8) tim btwn two succssiv arrivals (assumd units is hours) t) numbr of arrivals in t hours t) ~ Poisson( αt ) α, from problm statmnt Th probability of waiting 4 hours or lss is qual to minus its complmnt (waiting mor than 4 hours). 4) > 4) Th probability that th waiting tim is mor than 4 hours, is th sam as, th probability that zro arrivals occur in th nxt 4 four hours. > 4) t 4) ) Th probability can thn b dtrmind using th Poisson distribution. t 4) ) α t ()(4) 4 ( α t) t)! N ( t) ( 4)!.987 Answr: 4).987
5 ENGR 33 Bautiful Homwork #9 5 of 5 Johnston Problm 4.58 PART (d) SOLUTION tim btwn two succssiv arrivals (assumd units is hours) ~ Exp ( ) What is th probability that th tim btwn two succssiv arrivals is btwn two and fiv hours (inclusiv)? This probability can b solvd by intgrating th pdf: 5 x Or, th probability can b dtrmind using th cdf: F(5;) F(;) ( ()(5) ) ( ()() ) Answr: 5).86 Poisson Mthod tim btwn two succssiv arrivals (assumd units is hours) t) numbr of arrivals in t hours t) ~ Poisson( αt ) α, from problm statmnt Th probability that th tim btwn two succssiv arrivals is btwn and 5 hours can b dtrmind using th Poisson distribution in th sam way part (c) was solvd. First th xponntial probabilitis ar xprssd as thir complmnts. Thn th xponntial probabilitis ar convrtd into Poisson probabilitis. (For xampl, th probability of waiting mor than 5 hours, is th sam as, th probability that thr ar zro arrivals in th nxt 5 hours.) Last, th Poisson probability is calculatd. [ [ t 5) )] ( ()(5) 5 ) > 5)] ( 5)! ( ) [ > [ ) ).9933 t ) )] ()() ( )! Answr: 5).86
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