Assume that, during each second, a Dartmouth switchboard receives one call with probability .01 and no calls with probability .99. Use the Poisson approximation to estimate the probability that the operator will miss at most one call if she takes a 5-minute coffee break.
The Poisson distribution is indeed appropriate here, because we are seeking to understanding the likelihood of a certain number of discrete occurrences of a particular event, knowing the mean rate of occurrence for that (independent) event in a fixed time period. The distribution is given by the following formula…
\[ P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!} \] …in which \(\lambda\) is the mean probability over a given time period, \(X\) is any random amount of occurrences for a time period, and \(k\) is the specific number of occurrences of interest.
Here, because our mean rate \(\lambda\) and \(k\) number of occurrences use different time periods, we need to standardize. I believe the easiest way to do this is to determine a new \(\lambda\) for the expected number of calls in a 5-minute period. Since there are \(60\) seconds in a minute, there are \(60*5=300\) seconds in 5 minutes. At an expected value of \(1/100\) calls per minute, our \(\lambda\) for 5 minutes would therefore be \(300 * (1/100) = 3\).
Since there are two possible \(k\) values to satisfy the condition of “at most one phone call” in the given time period, we would simply take the probability of those two outcomes (\(k=0\) and \(k=1\)) and add them together.
\[ P(X = 0) = \frac{e^{-3}*3^0}{0!} \]
P0 = exp(1)^-3\[ P(X = 1) = \frac{e^{-3}*3^1}{1!} \]
P1 = exp(1)^-3*3\[ P(X\;is\;1\;or\;less) = P(0) + P(1) \]
P0 + P1## [1] 0.1991483The chance of the operator missing at most one call is 0.1991, or roughly 19.9%.