Suppose you call some hotline and you are the 56th person in line, excluding the person currently being served. Callers depart according to a Poisson process with a rate of 2 per minute. What is the probability you will have to wait for more than 30 minutes ?

There are at least 3 ways to approach this.

Poisson

$\quad\mathbf{P}$(take > 30 min):

$$\begin{aligned} \sum_{k=0}^{55} \mathbf{P}_{\lambda}(k, \tau) &= \sum_{k=0}^{55} \frac{(\lambda\tau)^k e^{-\lambda\tau}}{k!} \\ &= \sum_{k=0}^{55} \frac{(2\cdot 30)^ke^{-2\cdot 30}}{k!} \approx 0.285491 \end{aligned}$$

Erlang

$\quad 1 - \mathbf{P}$(take $\le$ 30 min):

$$\begin{aligned} 1 - \int_{0}^{30} \frac{\lambda^{k} t^{k-1} e^{-\lambda t}}{(k-1)!}\, dt\, &= 1 - \int_{0}^{30} \frac{2^{56} t^{55} e^{-2t}}{55!}\, dt \\ & \approx 1 - 0.714509 = 0.285491 \\ \end{aligned}$$

$\quad$See here.

CLT

$$\begin{aligned} \mathbf\mu &= \frac{n}{\lambda} = \frac{56}{2} = 28 \\ \sigma^2 &= \frac{n}{\lambda^2} = 14 \\ \mathbf{P}(T > 30) &= 1 - \mathbf{P}\left(\frac{30- \mu}{\sigma}\right) \\ &\approx 1 - \phi(0.5345) \approx 0.2981 \\ \end{aligned}$$

See Example 6.12 of Introduction to Probability, 2nd Edition.

# posted by rot13(Unafba Pune) @ 9:20 AM