Refer to Exercise $3.122 .$ Assume that arrivals occur according to a Poisson process with an average of seven per hour. What is the probability that exactly two customers arrive in the twohour period of time between a. 2: 00 P.M. and 4: 00 P.M. (one continuous two-hour period)? b. 1: 00 P.M. and 2: 00 p.M. or between 3: 00 p.M. and 4: 00 P.M. (two separate one-hour periods that total two hours)?
Added by Sherry F.
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Given: Average arrival rate (lambda) = 7 per hour The probability mass function of the Poisson distribution is: \[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \] For X = 2 (exactly two customers arriving): \[ P(X = 2) = \frac{e^{-7} 7^2}{2!} \] \[ P(X = 2) = Show more…
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