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Probability, Statistics and Queueing

Requests reach a server as Poisson arrivals (independent, random arrivals) at an average of 6 per minute. Which of the following correctly combines the distribution followed by the intervals between arrivals and their mean?

  1. AUniform distribution, mean 6 seconds
  2. BExponential distribution, mean 10 seconds
  3. CNormal distribution, mean 10 seconds
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Answer: B. Exponential distribution, mean 10 seconds

With Poisson arrivals, the intervals between arrivals follow an exponential distribution whose mean is 1/λ, the reciprocal of the arrival rate λ. With λ = 6 per minute, the mean interval is 1/6 minute = 10 seconds. The exponential distribution makes short intervals most likely, with a long gap now and then. The first M in the M/M/1 queueing model stands for arrivals of this kind.

  • A. The intervals of Poisson arrivals do not follow a uniform distribution, and the mean mistakes the arrival rate of 6 for seconds.
  • B. Correct. The intervals follow an exponential distribution with a mean of 60 seconds ÷ 6 = 10 seconds.
  • C. The mean of 10 seconds is right, but the intervals of Poisson arrivals follow an exponential distribution, not a normal distribution spread symmetrically around the mean.
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