Question 333801
A certain large city averages three murders per week and their occurrences follow a Poisson distribution.
a) what is the probability that there will be five or more murders in a given week?
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P(x>=5)=1-P(X<=4) = {{{1-sum(3^x*e^-3/x!)}}} summed from x=0 to x=4
={{{1-0.815=0.185}}}
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b)on the average, how many weeks a year can this city expect to have no murders?
{{{52*P(x=0)=52*(3^0*e^-3/0!)=52*e^-3=52*(0.0498)=2.59}}} or about 3 weeks
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c) how many weeks per year(on average) can the city expect the number of murder per week to equal or exceed the average number per week?
52*P(x>=3)=52*(1-P(x<=2)= {{{52*(1-sum( 3^x*e^-3/x!))}}} summed for x=0 to x=2
={{{52*(1-0.423)=29.99}}} or about 30 weeks