I'm reviewing for finals, and we were given some review questions without answers. It has been a while since we covered this topic, and I can't seem to find the method in my notes.
The problem statement:
Use formulas to carry out all computations in this problem
The number of calls received at a 24-hour service desk follows a Poisson distribution that has a standard deviation of 0.8 per minute.
(a) Find the probability that the time between the last call and the next call will be between 1 and 4 minutes
(b) Let X be the time until he receives his 3rd call. Find the $P(X \lt 5)$.
So I know that SD = 0.8, and the mean $\lambda$ = $0.8^2$ = $0.64$. For part (a), would I simple do $P(X=4) – P(X=1)$ using the pmf for a Poisson distribution? And for part (b), I believe there is a distribution that represents this type of problem. Would it be an exponential distribution? My notes aren't quite as clear as I had hoped.
Best Answer
(a) To get the probability that it happens between $1$ and $4$ minutes, you want the probability that it does not happen in the first minute but then occurs in the second minute plus the probability that it does not happen in the first $2$ minutes but then does it the third minute, etc.
We want the probability that it happens sometime in the interval $1-2$, $2-3$, or $3-4$.
For a given one minute interval we have,
$$P(\text{occurence)}=1-P(\text{does not occur)}=1-e^{-.64}\approx0.4727$$
Thus we get the following sum:
$$\sum_{n=1}^3 \left(e^{-.64}\right)^n \left(1-e^{-.64}\right)\approx 0.44999$$
(b) Note that this is the compliment of finding the probability that he gets $2$ or less calls in the first $5$ minutes. The mean for the $5$ minutes is $5\cdot0.64=3.2$
$$\frac{3.2^0e^{-3.2}}{0!}+\frac{3.2^1e^{-3.2}}{1!}+\frac{3.2^2e^{-3.2}}{2!}\approx0.3799$$
Thus we get
$$1-0.3799\approx0.6201$$