The Poisson Distribution is often used as a model for the number of events occuring within a specific time period.
The Poisson probability mass function is given by:
P(X=x) = (mu^x)*exp(-mu)/x! for x = 0,1,2...
where mu is the mean number of events in the given time period.
Example:
A toll-free phone number is available from 9AM to 9PM for your customers to register a complaint with a product purchased from your company. Past history indicates that an average of 0.4 calls are received per minute.
a. What properties must be true about the situation described above in order to use the Poisson distribution to calculate probabilities concerning the number of phone calls received in a one minute period?
The probability of receiving a phone call over a small interval is approximately proportional to the size of that interval.
The probability of receiving two calls in the same narrow interval is negligible.
The probability of receiving a phone call within a certain interval does not change over different intervals.
The probability of receiving a phone call in one interval is independent of the probability of an event in any other non-overlapping interval.
The last two assumptions are critical.
b. Zero phone calls will be receiver?
Let X be the number of phone calls.
X is distributed as Poisson with mean=0.4
P(X = 0)
=(0.4^0)*exp(-0.4)/0!
=0.6703
c. Three or more phone calls will be received?
P(X = 1)
=(0.4^1)*exp(-0.4)/1!
=0.2681
P(X = 2)
=(0.4^2)*exp(-0.4)/2!
=0.0536
P(X>=3)=1-P(X=0)-P(X=1)-P(X=2)
=1-0.6703-0.2681-0.0536
=0.0008
d. What is the maximum number of phone calls that will be received in a one minute period 99.99% of the time?
P(X=5)=(0.4^5)*exp(-0.4)/5!
=0.0001
Thus the maximum number of phone calls is 5.
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