# Call Center Poisson Process Question

Calls arrive at a call center according to a Poisson process with rate $$\lambda = 3$$ per minute.

Suppose there have been exactly 60 calls between 12pm and 12:30pm. Given this information, compute the probability of at least 60 calls between 12:30pm and 1pm.

I'm trying to understand the Poisson Process. Since events in disjoint time intervals are independent would it be $$P(X \ge 60 \mid Y = 60) = P(X \ge 60)*P(Y=60)/P(Y=60)$$? Am I taking the right approach or is there something I'm missing?

• First, note that the prior rate is irrelevant to calculating the number in your second period. That's the foundational idea of Poisson distributions. It is "memoryless." Oct 24, 2020 at 18:45
• It seems right to me, since $P(X\geq 60 \cap Y=60)=P(X\geq 60) \cdot P(Y=60)$ Oct 24, 2020 at 18:49
• You are on the right track, but you need to match the Poisson mean for $X$ in $P(X \ge 60)$ with the relevant 30 min time period. Oct 24, 2020 at 23:22

As @DavidG.Stork Comments, you can ignore what happened before 12:30, as long as you're not using that information to estimate $$\lambda.$$

Therefore, let $$X \sim \mathsf{Pois}(\lambda = 90),$$ where the rate $$\lambda = 90$$ = (30 min)(3/min). Then you want $$P(X \ge 60) = 0.99967.$$

You can get this exact answer using R, where ppois is a Poisson CDF, as shown below. Some statistical calculators could do essentially the same omputation.

1 - ppois(59, 90)
[1] 0.9996747


You might try a normal approximation to this Poisson distribution, $$\mathsf{Norm}(\mu = 90, \sigma=\sqrt{90}),$$ standardize, and use printed tables of CDF of standard normal to get a reasonable normal approximation (with continuity correction).

The normal approximation from R, where pnorm is a normal CDF, as shown below:

1 - pnorm(59.5, 90, sqrt(90))
[1] 0.9993477


Using normal tables you would get somewhat less accurate version of this approximation, because some rounding error is involved in using such a table.

The figure below, compares $$\mathsf{Pois}(\lambda=90),$$ centers of red circles, with the density function of $$\mathsf{Norm}(\mu=90, \sigma=\sqrt{90}).$$

R code for figure:

 curve(dnorm(x, 90, sqrt(90)), 0, 140, lwd=2, ylab="PDF", main="")
abline(v=0, col="green2")
abline(h=0, col="green2")
k = 0:140; pdf=dpois(k, 90)
points(k, pdf, col="red")
abline(v = 59.5, col="blue", lwd=2, lty="dotted")