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Problem 107: In a given city, 6% of all drivers get at least one parking ticket per year. Use the Poisson approximation to the binomial distribution to determine the probabilities that among 80 drivers randomly chosen in this given city:
a. 4 will get at least one parking ticket in any given year,
b. At least 3 will get at least at least one parking ticket in any given year,
c. Anywhere from 3 to 6, inclusive will get at least one parking ticket in any given year.

Problem 107: In a given city, 6% of all drivers get at least one parking ticket per year. Use the Poisson approximation to the binomial distribution to determine the probabilities that among 80 drivers randomly chosen in this given city:
a. 4 will get at least one parking ticket in any given year,
b. At least 3 will get at least at least one parking ticket in any given year,
c. Anywhere from 3 to 6, inclusive will get at least one parking ticket in any given year.


Solution:


Poisson distribution, $f(x;\lambda )=\frac{{\lambda }^{x}{e}^{-\lambda }}{x!}$

a. 4 will get at least one parking ticket in any given year
Given that $x=4, n=80, p=0.06, np = 4.8$
$f(4;4.8)=\frac{{4.8}^{4}{e}^{-4.8}}{4!} = 0.182029$


b. At least 3 will get at least at least one parking ticket in any given year
Given that $x\ge 3, n=80, p=0.06, np = 4.8$
$P(x\ge 3) = 1-P(x<3) = 1-[P(x=0)+P(x=1)+P(x=2)]$
$P(x=0) = f(0;4.8)=\frac{{4.8}^{0}{e}^{-4.8 }}{0!} = 0.00823$
$P(x=1) = f(1;4.8)=\frac{{4.8}^{1}{e}^{-4.8 }}{1!} = 0.039503$
$P(x=2) = f(2;4.8)=\frac{{4.8}^{2}{e}^{-4.8 }}{2!} = 0.094807$
$P(x\ge 3) = 1-[P(x=0)+P(x=1)+P(x=2)] = 1-(0.00823+0.039503+0.094807) = 1-0.1425 =0.857.$

c. Anywhere from 3 to 6, inclusive will get at least one parking ticket in any given year
Given that $3 \le x\le 6, n=80, p=0.06, np = 4.8$
$P(3 \le x\le 6) = P(x=3)+P(x=4)+P(x=5)+P(x=6)]$
$P(x=3 = f(3;4.8)=\frac{{4.8}^{0}{e}^{-4.8 }}{0!} = 0.151691$
$P(x=4) = f(4;4.8)=\frac{{4.8}^{1}{e}^{-4.8 }}{1!} = 0.182029$
$P(x=5) = f(5;4.8)=\frac{{4.8}^{3}{e}^{-4.8 }}{3!} = 0.174748$
$P(x=6) = f(6;4.8)=\frac{{4.8}^{4}{e}^{-4.8 }}{4!} = 0.139798$
$P(3 \le x\le 6) = P(x=3)+P(x=4)+P(x=5)+P(x=6)] = 0.162$

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