Answer to Question #348904 in Statistics and Probability for Michu

Question

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​

Answer 1

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

The Poisson distribution approximates the binomial distribution

$$\lambda=np=4.8$$

(1)

$$P(X=4)=\dfrac{e^{-4.8}(4.8)^4}{4!}=0.182029$$

(2)

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

$$-P(X=2)=1-\dfrac{e^{-4.8}(4.8)^0}{0!}$$

$$-\dfrac{e^{-4.8}(4.8)^1}{1!}-\dfrac{e^{-4.8}(4.8)^2}{2!}$$

$$=0.857461$$

(3)

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

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

$$=\dfrac{e^{-4.8}(4.8)^3}{3!}+\dfrac{e^{-4.8}(4.8)^4}{4!}$$

$$+\dfrac{e^{-4.8}(4.8)^5}{5!}+\dfrac{e^{-4.8}(4.8)^6}{6!}=0.648265$$