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Math Help - Can someone please help me with this?

  1. #1
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    Exclamation I need your help, PLEASE read!!

    I really need help on this, thank you in advance for your help, I appreciate it so much.

    In a warehouse, accidents have been taking place at the rate of 2 every 3 months.
    a) What probability distribution is most likely to model the number of accidents in a 3 month period (This is be selected from either the Binomial, Poisson or normal distribution)? Explain why you have made your choice and give any parameter(s) for the distribution.
    b) Find the mean and standard deviation of the number of accidents per year
    c) What is the percentage of months with no accidents?
    d) If a month is chosen at random, find the probability of at least one accident.
    Last edited by Kiwigirl; March 28th 2006 at 02:21 PM.
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  2. #2
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    Poisson distribution is usually used to model number of accidents. Binomial distribution is not good here because it takes only values from 0 to N (you must specify N but it is not specified in the problem). Normally distributed random variable can take negative values. As the parameter lambda of Poisson distribution is equal to the mean of the random variable with this Poisson distribution then lambda=2 for a 3-month period. For a year period lambda=2*4=8 and for a month period lambda=2/3.

    As we said before the parameter lambda of Poisson distribution was equal to the mean of the random variable with this Poisson distribution. The variance of this random variable is also equal to lambda. so for a year period mean=8 and standard deviation = SQR(variance)=2*SQR(2).

    Probability of the event that no accident will happen this month is equal to the probability that a random variable with Poisson distribution (lambda=2/3 as it is a month period) is equal to 0. So it is EXP(-lambda)=EXP(-2/3)=0,5134. So approximately 51,34% of months are without accidents.

    The probability of at least one accident for a month is 1 minus probability of the event that no accident happens this month. So it is equal to 1-0,5134=0,4866.
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  3. #3
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    Reply

    Wow, that sounds really complicated. Could you explain the answer of this question in a more simpler form for me to understand? Thank you
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  4. #4
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    Quote Originally Posted by Kiwigirl
    Wow, that sounds really complicated. Could you explain the answer of this question in a more simpler form for me to understand? Thank you
    Let's remember something from the theory of probabilities. If a random variable  \phi has Poisson distribution then
    <br />
P(\phi = k)=\frac{\lambda^k}{k!}exp(-\lambda)<br />
    where  \lambda is a parameter.
    So
    <br />
P(\phi = 0)=\frac{\lambda^0}{0!}exp(-\lambda)=exp(-\lambda)<br />
    (see the solution of c) )
    If we want to find a probability of the event that  \phi>0 (as in d) then
    <br />
P(\phi > 0)=1-P(\phi = 0)<br />
    It follows from a property of probabilities that the probability of the complementary event is equal to 1 minus probability of that event.
    The fact that the mean and variance of Poisson distribution are equal to  \lambda is a well-known property of Poisson distribution.
    Is everything clear now?
    If not please tell me what is unclear.
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