Math - Statistics

posted by .

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.

  • Math - Statistics -

    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

  • Math - Statistics -

    Oh i'm sorry:

    The check-out can process a total of four customers every minute.

  • Math - Statistics -

    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

  • Math - Statistics -

    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.

  • Math - Statistics -

    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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Statistics

    Hi folks, I'm having trouble with figuring out the intensity matrix for a particular birth/death process. Customers arrive in pairs in a Poisson stream with intensity lambda. There is waiting room for one customer. Service time is …
  2. statistics

    1. Observe the number of customers visiting a particular ATM Centre during any day from March 21 to 26, 2011 between 10 a.m. – 12 noon with a time interval of 5 minutes (number of customers using the ATM every 5 minutes). Calculate …
  3. Statistics

    Suppose cars arrive at Burger King's drive-through at the rate of 20 cars every hour between 12:00 noon and 1:00 pm. A random sample of 40 one-hour periods between 12:00 noon and 1:00 pm is selected and has 22.1 as the sample mean …
  4. Math/Statistics

    STATISTICS PLEASE HELP!!!!! I am sorry this is jy second time posting, but I really need help. I have tried contacting my teacher and classmates with no luck. Please just steer me in the right direction!: Suppose cars arrive at Burger …
  5. statistics

    3. At a cafeteria the customers arrive at an average of 0.3 per minute. The probability that a)exactly 2 customers arrive in a 10 minute span b)2 or more customers arrive in a 10 minute span c) one customer arrives in a 5 minute span …
  6. Statistics

    The number of traffic accidents per day on a certain section of highway is thought to be Poisson distributed with a mean equal 2.19. Based on this, how many traffic accidents should be expected during a week long period?
  7. Statistics

    The probability distribution of number of calls coming to a call center every 15 minute period is shown below. x 5 6 7 8 9 10 11 12 13 p ( x ) 0.02 0.04 0.07 0.08 0.09 0.11 0.12 0.13 0.10 Continued… x 14 15 16 17 18 19 20 p ( x ) …
  8. statistics

    15. A financial consultant conducts seminars with each seminar limited to 20 attendees. Because of the small size of the seminar, and the personal attention each person receives, a large number of attendees are expected to become clients. …
  9. statistics

    TRUE OR FALSE 35. A survey of women owned business found that approximately 30% of all small businesses are owned by non‐Hispanic white women. In a sample of 250 non‐Hispanic businesses owned by whites, the average number …
  10. statistics

    At a cafeteria the customers arrive at an average of 0.3 per minute. The probability that a)exactly 2 customers arrive in a 10 minute span b)2 or more customers arrive in a 10 minute span c) one customer arrives in a 5 minute span …

More Similar Questions