Question

Suppose the given country road has large defects which are distributed along its length according to Poisson distribution with an average of 2 defects per mile. It takes the crew repairing these defects one full day to fix one of them. They move along the road and repair the defects as they are found.

a) If the road is 10 miles long what is the probability they will be done in three weeks (15 working days)?

b) What is the probability they will get at least to the middle of the road (i.e. advance by at least 5 miles) by the end of the second week (after 10 working days)?

c) What is the mean and standard deviation of the length of the road they will cover by the end of the second week?

Answer 1

Solution:

Given that,

a) Let X be the number of defects in k miles

Average number of defects in k miles = 2k

Then X ~Poisson( = 2k)

For k = 10 miles,

X ~Poisson( = 20)

Probability they will be done in three weeks  (15 working days)

= P(X 15)

= 0.1565 (Using R command ppois(15,20) )

b) For k = 5 miles,

X ~Poisson( = 10)

Probability they will get at least to the middle of the road (i.e. advance by at least 5 miles) by the end of the second week (after 10 working days) = P(X 10) (If the defects are less or equal to 10 in 5 miles, crew will take at most 10 working days and will get at least to the middle of the road)

= 0.5830   (Using R command ppois(10,10) )

c) By end of second week (10 working days), 10 defects will be repaired.

X ~Poisson( = 10)

E(X) = = 10

Standard Deviation (X) == = 3.16

Number of miles = X / 2

Mean number of miles = E(X)/2 = 10 / 2 = 5 miles

Variance of number of miles = var(X/2) = Var(X)/4

Standard deviation of number of miles = SD(X) / 2 = 3.16 / 2 = 1.58 miles

Standard deviation of number of miles =1.58 miles