Let \(x\) be the number of callers put on hold. We want the probability that one or more people are put on hold to be no more than 1%: \[ \begin{aligned} P(x \geq 1) & \leq 0.01\\ 1-P(x=0) & \leq 0.01\\ -P(x=0) & \leq -0.99\\ P(x=0) & \geq 0.99\\ \end{aligned} \] Or in other words the probability that no people get put on hold is at least 99%.

Let \(k\) be the random number of 911 callers at any given minute. Since all 911 calls are exactly one minute long (i.e. you won’t have hour-long calls for example), to ensure none of the \(k\) callers is put on hold, the call center only needs to have at least \(k\) operators on hand at any given minute. Ex: If you have 2 operators and either 0, 1, or 2 people call, then no one gets put on hold.

Since the population of New York City is large and for any given citizen of New York to make a 911 call is a rare event, \(k\) can be modeled as a Poisson random variable with mean \(= \lambda=1.1\). For a range of values of \(k=0, 1, 2, \ldots\), we plug in \(k\) and \(\lambda=1.1\) into the formula on page 148: \[ P(\mbox{observe $k$ rare events}) = \frac{\lambda^k e^{-k}}{k!} \] Instead of doing this by hand, we can use the R command I gave in class: dpois(k, lambda=1.1)

k <- c(0, 1, 2, 3, 4, 5, 6)
prob <- dpois(k, lambda=1.1)
round(prob, 3)
## [1] 0.333 0.366 0.201 0.074 0.020 0.004 0.001

So the probability of getting exactly \(k\) callers is

k 0 1 2 3 4 5 6
Probability 0.333 0.366 0.201 0.074 0.020 0.004 0.001

and therefore the probability of \(k\) or less callers is

k or less 0 or less 1 or less 2 or less 3 or less 4 or less 5 or less 6 or less
Probability 0.333 0.699 0.900 0.974 0.995 0.999 1.000

Since 4 is the first value of \(k\) such that the probability exceeds 99%, we need to have at least 4 operators on hand to ensure that the probability that someone gets put on hold is no more than 1%.