Question

In 2011, New York City has a total of 11,232 motor vehicle accidents that occurred on Monday through Friday between the hours of 3 p.m. and 6 p.m. (new York State Department of Motor Vehicles website, October 24, 2012). This corresponds to mean of 14.4 accidents per hour.
a. Compute the probability of no accidents in a 15-minute period.

b. Compute the probability of at least one accident in a 15-minute period.

c. Compute the probability of four or more accidents in a 15-minute period.

Answer 1

Answer:

$a. \ P(X=0)=5.574\times10^{-7}$

$b. \ \ P(X\leq 1)=8.584\times10^{-6}$

$c.\ \ \ P(X\geq 4)=0.9997$

Step-by-step explanation:

a. #We notice this is a Poisson probability function expressed as:

$f(x)=\frac{\mu^xe^{-\mu}}{x!}$

x-number of occurrences in a given interval.

$\mu$-mean occurrences of the event

-The mean is calculated as:

$\mu=\frac{14.4}{4}=3.6$

#the probability of no accidents in a 15-minute period is :

$P(x)=\frac{\mu^xe^{-\mu}}{x!}\\\\P(X=0)=\frac{14.4^0e^{-14.4}}{0!}\\\\=5.574\times10^{-7}$

Hence, the probability of no accident in a 15-min period is $=5.574\times10^{-7}$

b. The the probability of at least one accident in a 15-minute period. is calculated as:

$P(x)=\frac{\mu^xe^{-\mu}}{x!}\\\\P(X\leq 1)=\frac{14.4^0e^{-14.4}}{0!}+\frac{14.4^1e^{-14.4}}{1!}\\\\=5.574\times10^{-7}+8.026\times 10^{-6}\\\\=8.584\times 10^{-6}$

Hence,  the probability of at least one accident in a 15-minute period is $8.584\times10^{-6}$

c. The probability of four or more accidents in a 15-minute period is calculated as:

$P(X\geq 4)=1-P(X\leq 3)=1-[P(X=0)+P(X1)+P(X=2)+P(X=3)]\\\\=1-[5.574\times10^{-7}+a. \ 8.026\times10^{-6}+\frac{14.4^2e^{-14.4}}{2!}+\frac{14.4^3e^{-14.4}}{3!}]\\\\=1-[8.584\times 10^{-6}+5.779\times10^{-5}+2.774\times10^{-4}]\\\\=0.9997$

Hence,the probability of four or more accidents in a 15-minute period. is  0.9997