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Example 6.18: Suppose that telephone calls arriving at a particular switchboard follow a Poisson process with an average of 5 calls coming per minute. What is the probability that up to a minute will elapse by the time 2 calls have come in to the switchboard? Solution: The Poisson process applies, with time until 2 Poisson events following a gamma distribution with ? = 1/5 and ? = 2. Denote by X the time in minutes that transpires before 2 calls come. The required probability is given by P(X ? 1) = ?_0^1 (1/?²)xe^(-x/?) dx = 25 ?_0^1 xe^(-5x) dx = 1 - e^-5(1 + 5) = 0.96. While the origin of the gamma distribution deals in time (or space) until the occurrence of ? Poisson events, there are many instances where a gamma distribution works very well even though there is no clear Poisson structure. This is particularly true for survival time problems in both engineering and biomedical applications.

          Example 6.18: Suppose that telephone calls arriving at a particular switchboard follow a Poisson process with an average of 5 calls coming per minute. What is the probability that up to a minute will elapse by the time 2 calls have come in to the switchboard? 
Solution: The Poisson process applies, with time until 2 Poisson events following a gamma distribution with ? = 1/5 and ? = 2. Denote by X the time in minutes that transpires before 2 calls come. The required probability is given by
P(X ? 1) = ?_0^1 (1/?²)xe^(-x/?) dx = 25 ?_0^1 xe^(-5x) dx = 1 - e^-5(1 + 5) = 0.96.
While the origin of the gamma distribution deals in time (or space) until the occurrence of ? Poisson events, there are many instances where a gamma distribution works very well even though there is no clear Poisson structure. This is particularly true for survival time problems in both engineering and biomedical applications.

Example 6.18: Suppose that telephone calls arriving at a particular switchboard follow a Poisson process with an average of 5 calls coming per minute. What is the probability that up to a minute will elapse by the time 2 calls have come in to the switchboard?
Solution: The Poisson process applies, with time until 2 Poisson events following a gamma distribution with ? = 1/5 and ? = 2. Denote by X the time in minutes that transpires before 2 calls come. The required probability is given by
P(X ? 1) = ?0^1 (1/?²)xe^(-x/?) dx = 25 ?0^1 xe^(-5x) dx = 1 - e^-5(1 + 5) = 0.96.
While the origin of the gamma distribution deals in time (or space) until the occurrence of ? Poisson events, there are many instances where a gamma distribution works very well even though there is no clear Poisson structure. This is particularly true for survival time problems in both engineering and biomedical applications.

Added by Silvia W.

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