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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
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