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    how would you do these Poisson problems

    accidents have been taking place at the rate of 2 every 3 months.
    a) assuming the accidents occur independently, calculate the mean and standard deviation of the number of accidents per year.
    b) if a month is chosen at random, calculate the probability of at least 1 accident.
    c) what is the percentage of months with no accidents?


    AND


    number of call-outs on weekdays for an ambulance has a poisson distreibution with 18 call-outs per 5-day week. the random variable X is the number of call-outs per weekday.
    a) calculate the mean, E(X)
    b) calculate the variance, VAR(X)
    c) calculate the standard deviation of X


    also iam wondering when do i normally use, normal distribution, poisson distribution and binomial distribution :eek:
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    (Original post by ronnie)
    how would you do these Poisson problems

    also iam wondering when do i normally use, normal distribution, poisson distribution and binomial distribution :eek:
    You might find this useful http://www.mathsnet.net/asa2/2004/s2.html#1
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    (Original post by ronnie)
    how would you do these Poisson problems

    accidents have been taking place at the rate of 2 every 3 months.
    a) assuming the accidents occur independently, calculate the mean and standard deviation of the number of accidents per year.
    b) if a month is chosen at random, calculate the probability of at least 1 accident.
    c) what is the percentage of months with no accidents?


    AND


    number of call-outs on weekdays for an ambulance has a poisson distreibution with 18 call-outs per 5-day week. the random variable X is the number of call-outs per weekday.
    a) calculate the mean, E(X)
    b) calculate the variance, VAR(X)
    c) calculate the standard deviation of X


    also iam wondering when do i normally use, normal distribution, poisson distribution and binomial distribution :eek:
    2 every 3 months => 8 every year ( So 8 is yearly mean).

    For Poisson distribution, mean = variance = 8.

    monthly mean = 2/3 .
    P[no accidents per month] = e^(-2/3).
    So P[at least 1] = 1 - e^(-2/3).

    On average, you would expect 1/3 of the months to have no accidents.
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    number of call-outs on weekdays for an ambulance has a poisson distreibution with 18 call-outs per 5-day week. the random variable X is the number of call-outs per weekday.
    a) calculate the mean, E(X)
    b) calculate the variance, VAR(X)
    c) calculate the standard deviation of X

    (a) average callouts per day = 18/5 =3.6
    (b) variance=mean=3.6
    (c) standard deviation = root 3.6= 1.897366596 blah blah blah

    I got a different answer to q1 part (c) though.
    X= number of accidents per month. X~Poi(2/3)
    Therefore P(X=0)= e^-2/3
    And percentage of random months with no accidents = 100e^-2/3, which =51.34...%
 
 
 
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