Question #348904

in a given city 6% of all drivers get at least one parking ticket per year . use the position approximation to the binomial distribution to determine the probabilities that among 80 drivers (randomly choosen in the city). (1) 4 will get at least one parking ticket in any given year, (2)at least 3 will get one parking ticket in any given year, (3)anywhere from 3 to 6 inclusive ,will get at least one parking ticket in any given year​

1
Expert's answer
2022-06-08T15:59:05-0400

n=80>20,np=80(0.06)=4.8<5n=80>20, np=80(0.06)=4.8<5

The Poisson distribution approximates the binomial distribution


λ=np=4.8\lambda=np=4.8

(1)


P(X=4)=e4.8(4.8)44!=0.182029P(X=4)=\dfrac{e^{-4.8}(4.8)^4}{4!}=0.182029

(2)


P(X3)=1P(X=0)P(X=1)P(X\ge3)=1-P(X=0)-P(X=1)

P(X=2)=1e4.8(4.8)00!-P(X=2)=1-\dfrac{e^{-4.8}(4.8)^0}{0!}

e4.8(4.8)11!e4.8(4.8)22!-\dfrac{e^{-4.8}(4.8)^1}{1!}-\dfrac{e^{-4.8}(4.8)^2}{2!}

=0.857461=0.857461

(3)


P(3X6)=P(X=3)+P(X=4)P(3\le X\le6)=P(X=3)+P(X=4)

+P(X=5)+P(X=6)+P(X=5)+P(X=6)

=e4.8(4.8)33!+e4.8(4.8)44!=\dfrac{e^{-4.8}(4.8)^3}{3!}+\dfrac{e^{-4.8}(4.8)^4}{4!}

+e4.8(4.8)55!+e4.8(4.8)66!=0.648265+\dfrac{e^{-4.8}(4.8)^5}{5!}+\dfrac{e^{-4.8}(4.8)^6}{6!}=0.648265


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