Firm has complaints, probability Complaints about an Internet brokerage firm occur at a rate of 5 per day. The number of complaints appears to be Poisson distributed.
A. Find the probability that the firm receives 3 or more complaints in a day.
B Find the probability that the firm receives 21 or more complaints in a 5-day period.
I got part A, using the Poison distribution, and doing 1-P(0,1 or 2) complaints. For part B, do I have to do 1-P(0 through 20) with the rate being 20 in a 5 day period. Seems pretty tedious. Should I use normal approximation instead?
 A: If $X_i$ are independent and distributed as Poisson($\lambda_i$) then $\sum_{i=1}^n X_i$ is Poisson($\sum_{i=1}^n \lambda_i$). This is a special property that the Poisson distribution shares with a few other important distributions, including the normal distribution.
This implies that the sum in your problem is distributed as Poisson($25$). So, this calculation would require adding up $21$ terms in principle, except most of them are very nearly zero. Indeed, I find that the sum of all the terms is about 0.1855, while the sum of only the last ten terms is about 0.1849.
A: I wonder if the instructor/text really wants you to do the simple,
but somewhat tedious math to compute and add so many Poisson probabilities. Possible options:
(a) Maybe your text has a table of the Poisson distribution, and
maybe $\lambda = 25$ is one of the entries.
(b) For $\lambda$ as large as 25, the normal approximation is pretty
good. Here $P(Y < 20.5) = 0.184,$ for $Y \sim Norm(\mu = 25, \sigma = 5).$ Look in your text for instructions about such approximations. (What are the mean and standard deviation of your
Poisson random variable?)
(c) If your course uses software, you can get the answer mentioned
by @Ian with minimal trouble. For example, in R statistical
software the statement 'ppois(20, 15)' returns the desired answer.
It would be rare to find a statistical computer package that does
not compute Poisson probabilities.
