We now give an example in which Poisson probabilities are found.

Example: Arrival of phone calls

A homogeneous Poisson process might be used to model arrival of telephone calls in an office. We will assume that calls arrive at a constant rate of \(\lambda = 3\) per hour during the morning (8:30 am to 12:30 pm).

  1. What are the mean and standard deviation of the number of calls arriving in a single hour?
  2. What is the probability of getting three or more calls in a single hour?
  3. What is the probability of getting no calls in a 20-minute period?

If \(X\) is the number of calls in an hour,

\[ X \;\; \sim \; \; \PoissonDistn(\lambda t = 3) \]

The mean and variance of \(X\) are both \(\lambda = 3\), so its standard deviation is \(\sqrt 3 = 1.73\).

The probability of 3 or more calls in an hour is

\[ \begin{align} P(X \ge 3)\;\; &= \;\; p(3) + p(4) + p(5) + \cdots \\[0.3em] &= \;\; 1 - \big(p(0) + p(1) + p(2) \big) \\[0.3em] &= \;\; 1 - \left(\frac {3^0 e^{-3}} {0!} + \frac {3^1 e^{-3}} {1!} + \frac {3^2 e^{-3}} {2!} \right) \\ &= \;\; 1 - (0.050 + 0.149 + 0.224 ) \\[0.3em] &= \;\; 0.577 \end{align} \]

If we use \(Y\) to denote the number of calls in a 20-minute period,

\[ Y \;\; \sim \; \; \PoissonDistn(\lambda t = 1) \]

The probability of zero calls in 20 minutes is therefore

\[ P(Y = 0) \;\; = \; \; \frac {1^0 e^{-1}} {0!} \;\; = \; \; 0.368 \]