Question

Suppose that potholes along the motorway north of Las Angles have been found to occur according to a Poisson process with rate parameter λ.

(a) Out of 46 three kilometer stretches of motorway, the average number of potholes was found to be 6.2. Find the maximum likelihood estimate for the rate parameter, λˆ MLE, and the standard deviation of the estimate, SD(λˆ MLE).

(b) Using the maximum likelihood estimate as the rate parameter, what is the probability that a randomly chosen three kilometer stretch of motorway has more than five potholes?

(c) If three randomly chosen, nonoverlapping, stretches of the freeway are checked by inspectors, what is the probability that only one of them has more than five potholes?

In parts (b) and (c), please identify the relevant variable and its distribution, and express the required probability in terms of this variable before calculating your answer. You may use the formula, tables or a calculator to find the actual probability. For example (in a different scenario) you might write: “Let N be the number of dry wells before drilling a well that finds oil, then N ∼ NB(n = 1, p = 0.25). We want Pr(N < 5) = ....”

Answer 1

Let the random variable X is defined as

X : Number of potholes in three kilometer stretches motorway.

X ~ Poisson (lamda)

From the information

n = 46 and Xbar = 6.2

a) MLE for lamda is the solution of likelihood equation

where

Taking log of both sides we get

-----------------(I)

Hence is an MLE for lamda

Since X ~ P(lamda) for Poisson distribution mean and variance are equal.

Mean(X) =Var(X) =lamda

b) X ~ Poisson(lamda = 6.2)

Required Probability = P ( X > 5)

= 1 - P (X < =5)

= 1 - F(5)

By using Excel Function

F(5) = POISSON(5,6.2,1) = 0.4141

P ( X > 5)= 1- 0.4141 = 0.5859

P ( Three kilometer stretch of motorway has more than 5 potholes) = 0.5859

c) Let X,Y and Z are three randomly chosen non overlapping streches of freeway.

X ~ Poisson (lamda = 6.2) , Y ~ Poisson ( lamda = 6.2) and Z ~ Poisson ( lamda = 6.2)

P ( X > 5) = 0.5859 , P (Z > 5) = 0.5859 and P ( Z > 5) = 0.5859

P ( Exactly one of them has more than 5 potholes) = P ( X has more than 5 potholes) or P ( Y hs more than 5 potholes) or P ( Z has more than 5 potholes)

= P ( X > 5) + P ( Y >5) + P (Z > 5 )

= 3*0.5859

= 1.7577

Since probability is never greater than 1

Hence P ( Exactly one of them has more than 5 potholes) = 1