1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Poisson Distrobution-HELP

  1. Jun 17, 2005 #1
    Poisson Distrobution-HELP!! URGENT!!!

    the problem:

    The Breakdowns of a robot follow a Poisson Dist. with an avg of .5 breakdowns per 8-hour workday. If this robot is placed in service at the beginning of the day, find the probability that:

    a. It will break down durring the day.
    b. It will work for at least 4 hours without breakdown
    c. Does what happend the day before have any effect on your answers? why?

    I'm a newb to probability and am I'm haveing trouble figuring out where to begin. I know that the rate(lamda) has something to do with the percent of break downs and the work day. Is it .5*workday or just .5. I just don't know where to begin with this problem. :cry:
     
  2. jcsd
  3. Jun 17, 2005 #2
    i know that P{X >= a } = e^(-גa) where ג is lamda, would P{ X < a } = -e^(-גa)?
     
  4. Jun 17, 2005 #3
    Well the answer to a. is

    P{ X < a } = e^(-גa)
    = e^(-0.5)
    = 0.6065

    where ג = 0.5 <- the rate equals lamda

    Since P{ X >= a } = 1 - ( 1 - e^( גa ) ) = e^(-גa)

    right?
     
  5. Jun 17, 2005 #4
    i need help on b
     
  6. Jun 18, 2005 #5
    anybody.......
     
  7. Jun 23, 2005 #6
    Answer


    A Poisson Process is a counting process; in your case define a success or occurrence when a breakdown happends. The probability of having n breakdowns within a period of time t is: P{N(t)=n}=exp(-lamda*t)*(lamda*t)^n/n!
    In this case lamda = 0.5 breakdowns/day (since a day has 8 hours of work. Also, we can express lamda per hour as lamda = 0.5/8 = 1/16 breakdowns/hour.

    a. It will break down during the day.
    Here we can look for the probability of having a breakdown in a day (8 hours period), then P{N(1) >= 1} = 1 - P{N(1) < 1} = 1 - P{N(1) = 0} = 1 - exp(-0.5)
    Or if we look for the probability that the first breakdown occurs within 8 hours; let say X1 is a random variable representing the interarrival time or interarrival occurrence time of breakdowns. By definition the interarrival times are ~Exponential, f(t) = lamda*exp(-lamda*t), 0 <= t < infinity
    Then P{X1 <= 8} = 1 - exp[-(0.5/8)*8] = 1 - exp(-0.5)

    b. It will work for at least 4 hours without breakdown.
    Here we look for the probability of no breakdown in at least 4 hours, or a breakdown in at most 4 hours.
    P{N(4) = 0} = exp[-(0.5/8)*4]
    P{X1 > 4} = 1 - P{X1 <= 4} = exp[-(0.5/8)*4]

    I hope it helps.
     
    Last edited: Jun 23, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Poisson Distrobution-HELP
  1. Help Poisson Errors. (Replies: 1)

Loading...