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The special case of the gamma distribution in which a is a positive integer n is called an Erlang distribution. If we replace B by - in the expression below, x 20 f(x; a, B) = *r(@) otherwise the Erlang pdf is as follows. F(x; 2, n) =- x20 (n - 1)! O x<0 It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter 2, then the total time X that elapses before all of the next n events occu (a) What is the expected value of X? E(X) = If the time (in minutes) between arrivals of successive customers is exponentially distributed with à = 0.5, how much time can be expected to elapse before the ninth customer arrives?

The special case of the gamma distribution in which a is a positive integer n is called an Erlang distribution. If we replace B by - in the expression below, x 20 f(x; a, B) = *r(@) otherwise the Erlang pdf is as follows. F(x; 2, n) =- x20 (n - 1)! O x<0 It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter 2, then the total time X that elapses before all of the next n events occu (a) What is the expected value of X? E(X) = If the time (in minutes) between arrivals of successive customers is exponentially distributed with à = 0.5, how much time can be expected to elapse before the ninth customer arrives?

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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The special case of the gamma distribution in which a is a positive integer n is called an Erlang distribution. If we replace B by 1 in the expression below, xa - 1-x/8 x 2 0 f(x; a, B) = { B®T(a) otherwise the Erlang pdf is as follows. (2(2x)" - le-ax (n - 1)! x2 0 f(x; A, n) = It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter a, then the total time X that elapses before all of the next n events occur has pdf f(x; A, n). (a) What is the expected value of X? E(X) = If the time (in minutes) between arrivals of successive customers is exponentially distributed with 1 = 0.5, how much time can be expected to elapse before the ninth customer arrives? min (b) If customer interarrival time is exponentially distributed with 2 = 0.5, what is the probability that the ninth customer (after the one who has just arrived) will arrive within the next 26 min? (c) The event {X < t} occurs if at least n events occur in the next t units of time. Use the fact that the number of events occurring in an interval of length t has a Poisson distribution with parameter At to write an expression (involving Poisson probabilities) for the Erlang cdf F(t; , n) = P(X s t). n- O F(t; 2, n) = 1 - k = 0 k! O F(t; 2, n) = 5 a)k k! k = 0 O F(t; 2, n) = 'ecae}k – 1 (k - 1)! k = 0 O F(t; 2, n) = 1 - k! k = 0 O F(t; 2, n) = 1- (k - 1)! k = 0
Transcribed Image Text:The special case of the gamma distribution in which a is a positive integer n is called an Erlang distribution. If we replace B by 1 in the expression below, xa - 1-x/8 x 2 0 f(x; a, B) = { B®T(a) otherwise the Erlang pdf is as follows. (2(2x)" - le-ax (n - 1)! x2 0 f(x; A, n) = It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter a, then the total time X that elapses before all of the next n events occur has pdf f(x; A, n). (a) What is the expected value of X? E(X) = If the time (in minutes) between arrivals of successive customers is exponentially distributed with 1 = 0.5, how much time can be expected to elapse before the ninth customer arrives? min (b) If customer interarrival time is exponentially distributed with 2 = 0.5, what is the probability that the ninth customer (after the one who has just arrived) will arrive within the next 26 min? (c) The event {X < t} occurs if at least n events occur in the next t units of time. Use the fact that the number of events occurring in an interval of length t has a Poisson distribution with parameter At to write an expression (involving Poisson probabilities) for the Erlang cdf F(t; , n) = P(X s t). n- O F(t; 2, n) = 1 - k = 0 k! O F(t; 2, n) = 5 a)k k! k = 0 O F(t; 2, n) = 'ecae}k – 1 (k - 1)! k = 0 O F(t; 2, n) = 1 - k! k = 0 O F(t; 2, n) = 1- (k - 1)! k = 0
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