Question : In a road traffic survey the

Krish Crosby

Krish Crosby

Answered question

2022-09-26

Question : In a road traffic survey the number of cars passing a marker at a road was counted. The streams of cars in the two directions were a priori modelled as independent Poisson processes of intensities 2 per minute and 3 per minute, respectively. It was decided to stop the counting once 400 cars had passed. Let T 400 be that (random) time point and compute, using appropriate approximations, a time α such that P ( T 400 α ) = 0.90
Claimed Answer 85.1 minutes.
My Attempt : Since 400 cars passing with a rate of 5 cars/minute to pass in approximately 400 5 = 80 min. However this is not correct.

Answer & Explanation

Nancy Phillips

Nancy Phillips

Beginner2022-09-27Added 12 answers

Let Xi be the time between the i-th and (i−1)-th arrival. Since the sum of two independent Poisson processes in again a poisson process, we know that X i E x p ( 1 / 5 )ю Furthermore, we know that the sum of n identical and independent exponential random variables with mean 1 / λ is E r l a n g ( n , λ ) distributed.
Using this we get that
P ( T 400 < a ) = P ( i = 1 400 X i < a ) = P ( E r l a n g ( 400 , 1 / 5 ) < a ) = 0.9.
With the R code qgamma(0.9, shape = 400, scale = 1/50) we get that this is the case for a = 85.16712
Edit . My first answer was an exact result. To get an approximate result we can use the central limit theorem. We have that
P ( i = 1 400 X i < a ) = P ( i = 1 400 400 1 5 400 1 5 2 < a 400 1 5 400 1 5 2 ) = P ( Z < a ) ,
where Z is a standard normal random variable and a = a 400 1 5 400 1 5 You can look Φ ( 0.9 ) up in a table and solve a = Φ 1 ( 0.9 ) for a to get 85.1 as

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Research Methodology

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?