Question

Alt Ctri (20 points) On one particular road in the city, th follows a Poisson distribution with ?-3, and the follows an exponential distribution with 0-8 hours. e number of traffic accidents time in between traffic accidents 5. A In a given day, what is the probablity that there are more than 5 traffic accidents in the city? B. What is the probability that another accident occurs within 5 hours since last C. one in that day? If 50 traffic accidents were recorded during a period of time. For each accident, the length of time since the previous accident to this one was recorded. What is the probability that the average of these 50 length of time is less than 5 hours?

PLEASE DO ASAP

Answer 1

Solution:-

1. Solu:-A)

Accident mean rate ? (lambda) = 3 accidents occur in inter arrival time t=8 hour

So, accident/hour will be ?t =3/8 accident’s /hour

So probability of 5 or more accident would occur

P(X ? 5) = P(X = 5) + P(X = 6) + . . . = 1 ? P(X < 5) = = 1 – {P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)}

Formula for

P(x,t)= {? (lambda) *t)^x* e-⁽? ?t⁾  } /x!

P(X=0) = {3/8)⁰e^-(3/8)} / 0! =.687

P(X=1) = {3/8)¹ e^-(3/8)} / 1! =.257

P(X=2)= {3/8)² e^-(3/8)} / 2! = {(.375)^2 *.687} /2 =0.0483

P(X=3)= 0.0060,  P(X=4)=0.00056

P(X ? 5) =1 – (.687+.257+.0483+0.0060+.00056) =1- (.9988) = 0.0012 Ans:-

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Solution:- B).

Find P(T ? 13|T > 8) means last accident occurred 8 hour before and we have to calculate probability next 5 hour ie 8+5=13 ,

From the Markov property

P(T ? 13|T > 8) = P(T ? 5)

So, P(T ? 5) = 1 ? e ?5? =1 - e^(-5*(3/8)) = 1- 0.1533 = 0.8467 Ans:--