Answer:
A. 0.1111
B. 0.1708 and 0
C. 0.338014.
Explanation:
Requirement A:
A discrete random variable X is said to have a Poisson distribution if its probability function is given by,
f(x ; λ)= (e^(-λ) * λ^x)/x! ; x= 0, 1, 2………..∞
where, e = 2.71828 and λ is the parameter of the distribution which is the mean number of success.
Given,
Mean number of accident λ = 4.5 per week
Let, X be the number of car accident which follow Poisson distribution. The probability of no accident occur in one week is,
P[no accident] = P[X=0]
= (e^(-4.5) *〖4.5〗^0)/0!
= 0.1111
The probability of no accident is 0.1111.
Requirement B:
Given,
Mean number of accident λ = 4.5 per week
Let, X be the number of car accident which follow Poisson distribution. The probability of 5 accidents occurs in one week is,
P[exactly 5 accidents] = P[X=5]
= (e^(-4.5) 〖4.5〗^5)/5!
= 0.1708
The probability of no accident is 0.1708.
And, The probability of more than 5 accidents occurs in one week is,
P[more than 5 accidents] = P[X>5]
=1- P[X is less than5]
= 1- P[X=0] - P[X=1] - P[X=2] - P[X=3] - P[X=4] - P[X=5]
=1 - 0.1111 - 0.4999 - 0.11248 - 0.1687 - 0.1898 - 0.17083
= -.2521
= 0
The probability of more than 5 accidents is 0.
Requirement C:
Poisson distribution is related to the experimental results expressed in two ways: successful and failure under some limiting condition.
Given,
Mean number of accident λ = 4.5 per week
= 4.5/7 per day
=.6429
Let, X be the number of car accident which follow Poisson distribution. The probability of one accident occurs in one day is,
P[exactly 1 accident] = P[X=1]
= (e^(-.6429) 〖.6429〗^1)/1!
= 0.338015
The probability of no accident is 0.338015.