Your problem needs to say something about how quickly people are processed at the checkout counter. Is the line supposed to be getting longer and longer while no one is checked out?
It also is not clear in (a) whether you are supposed to use the Poisson distribution of your string of random numbers.
Maybe Reiny or Damon can figure out the intent here
Suppose that the mean number of customers who arrive at the check-out counters each minute is 4. Create a Poisson Distribution with u (lamda) = 4 for x = 0 to 20. Compare your results to the histogram.
20 random numbers are generated with a Poisson distribution for u (lamda) = 4. Let the random number represent the number of arrivals at the check-out counter each minute for 20 minutes.
3, 3, 3, 3, 5, 5, 6, 7, 3, 6, 3, 5, 6, 3, 4, 6, 2, 2, 4, 1
a) How many customers were waiting after 5 minutes? 6 minutes? 7 minutes? 8 minutes?
b) Create a table that shows the number of customers waiting at the end of 1 through 20 minutes.
5 answers
Oh i'm sorry:
The check-out can process a total of four customers every minute.
The check-out can process a total of four customers every minute.
I must admit that I am not familiar with a Poisson distribution.
I googled it and checked several of the webpages.
For me to learn it and then explain it to you would be quite time-consuming, since you could surely learn it just as fast as this old brain of mine.
This Wolfram page looks quite complicated
http://mathworld.wolfram.com/PoissonDistribution.html
I googled it and checked several of the webpages.
For me to learn it and then explain it to you would be quite time-consuming, since you could surely learn it just as fast as this old brain of mine.
This Wolfram page looks quite complicated
http://mathworld.wolfram.com/PoissonDistribution.html
OK, now I think I see that they are asking for. A Poisson distribution tells you the probability of different numbers of "counts" that are expected in a particular time interval, when the average number per unit time is known. They must have already used a Poisson random distribution formula to get that series of numbers. Since you already have a set of numbers, you don't need to know the Poisson formula to answer their question.
After minutes one through four, there is no line. Fewer people arrive each minute than can be processed.
After minute five, there is one waiting, since 5 arrived and 4 were processed.
After minute six, 2 are waiting. During minute seven, 6 arrive and two are processed, so 4 are waiting after seven seconds. After minute eight,7 are waiting. Continue in this manner until you have completed a table for 20 seconds.
If you used the Poisson random number generator a second time, you'd get a different table. What is of most interest in such "queueing" problems is really the probability distribution of line length, but they did not ask you that.
After minutes one through four, there is no line. Fewer people arrive each minute than can be processed.
After minute five, there is one waiting, since 5 arrived and 4 were processed.
After minute six, 2 are waiting. During minute seven, 6 arrive and two are processed, so 4 are waiting after seven seconds. After minute eight,7 are waiting. Continue in this manner until you have completed a table for 20 seconds.
If you used the Poisson random number generator a second time, you'd get a different table. What is of most interest in such "queueing" problems is really the probability distribution of line length, but they did not ask you that.
known average rate = L (lambda you called it u) which is 4/min
or
L = 4
we want probability of 0, 1 , 2,3,4 arrivals per period
call each of these numers k
then Poisson dist means:
f(k,L) = L^k e^(-L) / k!
then generate with your calculator
eg
f(0.L) = 4^0 e^-4 / 0! =.0183
f(1,L) = 4^1 e^-4 / 1! =.1465
f(2,L) = 4^2 e^-4 / 2! =.1954
....
f(5,4) = 4^5 e^-4 /5! =.1563
f(6,4) = 4^6 e^-4/6! =.1042
graph that and compare. Notice peak around 3 and 4 arrivals per minute, the mean
or
L = 4
we want probability of 0, 1 , 2,3,4 arrivals per period
call each of these numers k
then Poisson dist means:
f(k,L) = L^k e^(-L) / k!
then generate with your calculator
eg
f(0.L) = 4^0 e^-4 / 0! =.0183
f(1,L) = 4^1 e^-4 / 1! =.1465
f(2,L) = 4^2 e^-4 / 2! =.1954
....
f(5,4) = 4^5 e^-4 /5! =.1563
f(6,4) = 4^6 e^-4/6! =.1042
graph that and compare. Notice peak around 3 and 4 arrivals per minute, the mean
1 answer