Probability of phone calls with exponential interarrival times

Question

Calls to a university’s admission information line are assumed to follow an exponential distribution with a mean of 8 minutes between each call.

1. What is the probability of there being no calls in a given hour?
2. What is the probability of at least 7 calls in a given hour?

I know the exponential distribution formula is $\lambda e^{-\lambda x}$. However I am now confused what to do next.

Answer 1

Let $N(t)$ denote the number of calls made to the information line by time $t$. The interperiod times between calls is (i.i.d.) exponentially distributed with mean 8. Let $\lambda$ denote the rate of the exponential distribution. Since $\lambda^{-1} = 8$, it follows that $\lambda = \frac{1}{8}$.

$N(t)$ is a counting process, but we can say more. Since the interperiod times between "arrivals" is iid exponential with rate $\lambda = \frac{1}{8}$, then $N(t)$ is a Poisson process. In particular, this means that $N(t)$ is distributed according to the distribution $\mathsf{Poisson}(\frac{t}{8})$.

1. The probability of there being no calls in a given hour is given by $P(N(60) = 0)$. The random variable $N(60)$ is distributed according to $\mathsf{Poisson}(7.5)$, so $P(N(60)= 0) = \exp(-7.5) \cdot 7.5^0 / 0! \approx 0.00055$.
2. The probability of at least 7 calls in a given hour is $P(N(60) \geq 7) = 1 - P(N(60) \leq 6) = 1- \exp(-7.5) \sum_{k=0}^{6} 7.5^k / k! \approx 0.62$.

Incidentally, the reason why we can reduce everything to $N(60)$ and not have to worry about terms like $N(60+t) - N(t)$ is because of the memorylessness property of the Poisson process.

Answer 2

Number of calls is Poisson. With calls arriving at the rate of one every $8$ minutes, that would be an average of $\lambda = 60/8 = 7.5$ per hour. As in @DavidKraemer's Answer (+1), The number of calls arriving in a given time interval is Poisson with the appropriate rate. [That answer makes a number of important points; I hope you will click to Accept it.]

No calls in an hour. For no calls in an hour, the probability from R (where ppois is a Poisson CDF) is $0.00055.$

ppois(0, 60/8)
[1] 0.0005530844

Using Exponential distribution: Another way to look at the problem is to let $W \sim \mathsf{Exp}(\mathrm{mean}=8; \mathrm{rate}=1/8)$, with $E(W) = 8$ min. mean wait, and to find the probability of waiting more than 60 minutes for the first call: $P(W > 60) = 1 - P(W \le 60).$

1 - pexp(60, 1/8)
[1] 0.0005530844

At least 7 in an hour. And the probability $P(X \ge 7) = 1-P(X \le 6) = 0.6218,$ for $X \sim \mathsf{Pois}(\lambda=60/8).$

1 - ppois(6, 60/8)
[1] 0.6218453

Using gamma distribution: One can use moment generating functions to show that the total waiting time (in minutes) $T_7$ for seven calls has $T_7 \sim \mathsf{Gamma}(\mathrm{shape}= 7, \mathrm{rate} = 1/8).$ Then the probability of at least seven calls in a hour is $P(T_7 \le 60) = 0.6218.$

pgamma(60, 7, 1/8)
[1] 0.6218453

Graphs. Here is a graph of the PDF of $\mathsf{Pois}(\lambda = 7.5).$

k = 0:30;  pdf = dpois(k, 60/8)
plot(k, pdf, type="h", lwd=4, main="PDF of POIS(7.5)")
abline(h=0, col="green2")
abline(v=0, col="green2")

Here are graphs of the PDFs of $\mathsf{Exp}(\mathsf{rate}=1/8)$ and $\mathsf{Gamma}(7. 1/8).$

par(mfrow=c(1,2))
 curve(dexp(x,1/8), 0, 60, lwd=2, ylab="PDF", 
       xlab="w", main="EXP(1/8)")
  abline(h=0,col="green2")
  abline(v=0,col="green2")
 curve(dgamma(x,7,1/8), 0, 200, lwd=2, ylab="PDF", 
      xlab="t", main="Gamma(7, 1/8)")
  abline(h=0,col="green2")
  abline(v=0,col="green2")
par(mfrow=c(1,1))