#### Example 2.1.17.

Suppose we observe traffic along a particular stretch of highway which typically has 200 cars pass per hour. Let \(N\) be the number of cars seen in a 2-hour observation period. Since the parameter \(\lambda = 200\) cars per hour is given for a 1-hour period, we should scale it to find the rate for a 2-hour period: \(\lambda t = (200)(2) = 400\) cars per 2-hours. Then, for example:
\begin{gather*} \Pr(N = 375) = p(375; 200, 2) = \frac{400^{375}}{375!} e^{-400} \approx 0.01. \end{gather*}
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Now consider the random variable \(M\) counting the number of cars seen in a 30-minute observation period. We should not calculate \(\lambda t = (200)(30) = 6000\text{,}\) since \(\lambda\) measures time in hours and \(t\) measures time in minutes. We should first convert \(t = 0.5\) hours, then we can find our appropriately scaled rate information: \((200)(0.5) = 100\) cars per half-hour. Then, for example:
\begin{gather*} \Pr(M = 110) = p(110; 200, 0.5) = \frac{100^{110}}{110!}e^{-100} \approx 0.02. \end{gather*}
Note that when we write the notation \(p(k; \lambda, t)\text{,}\) we assume that \(\lambda\) and \(t\) are expressed with the same time units already.
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