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?
 A: 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")

