Question 1028732: Potholes requiring repair in a section of a national highway occur at an average rate of 3.2 potholes per kilometre.
1. What is the probability that there are no potholes that required repair in 5km of the highway?
2. What is the probability that at most three potholes require repair in 200m?
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
Potholes requiring repair in a section of a national highway occur at an average rate of 3.2 potholes per kilometre.
1. What is the probability that there are no potholes that required repair in 5km of the highway?
2. What is the probability that at most three potholes require repair in 200m?
Solution:
In probability problems, recognizing an appropriate distribution is half the problem solved. The remaining part is just plugging in the parameters followed by a few key strokes on the calculator.
We will examine the problem with respect to the criteria for the Poisson distribution, as follows, extracted from
http://stattrek.com/probability-distributions/poisson.aspx
A Poisson experiment is a statistical experiment that has the following properties:
The experiment results in outcomes that can be classified as successes or failures (i.e. potholes or no potholes).
The average number of successes (μ) that occurs in a specified region is known. (yes, μ=3.2 potholes per km)
The probability that a success will occur is proportional to the size of the region (by the context of the problem, the longer the road, more potholes will be found)
The probability that a success will occur in an extremely small region is virtually zero. (chances of finding a pothole in a given metre length of road is very small).
Note that the specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc.
Thus the problem satisfies all the requirements of the Poisson distribution.
We can then calculate the required values accordingly.
The probability is given by:
P(x; μ) = (e^-μ) (μ^x) / x!
x=given number of occurrences
μ=average rate (=3.2 potholes/km)
e=natural log base = 2.7182818284...
n! = factorial function = n*(n-1)*....3*2*1
1. no potholes in 5 km.
μ for 5 km = 5*3.2=16
P(0;16)=e^(-16)*16^0/0!=e^(-16)*1/1=1.125*10^(-7)
2. up to 3 potholes in 200m
μ=3.2*(200/1000)=0.64 potholes / 200 m
P(x≤3; 0.64)
=P(x=0;0.64)+P(x=1;0.64)+P(x=2;0.64)+P(x=3;0.64)
=0.52729+0.33747+0.10799+0.02304
=0.99579
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