MATH
GROUP-3-WEEK_5.pdf

# GROUP-3-WEEK_5.pdf - Study Diary Week 5 Group 3 1 Exercise...

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Study Diary Week 5 Group 3 March 24, 2018 1 Exercise 1: The loss due to a fire in a commercial building is modeled by a random variable X with density function f ( x ) = ( 0 . 005(20 - x ) , 0 < x < 20 0 , otherwise Given that a fire loss exceeds 8, calculate the probability that it exceeds 16. Solution: P ( X > x ) = Z 20 x 0 . 005(20 - t ) dx = 0 . 005(200 - 20 t + 1 / 2 t 2 ) For 0 < x < 20 We have P ( X > 16 | X > 8) = ( P ( x > 16 , x > 8)) P ( x > 8) = P ( x > 16) P ( x > 8) = 1 9 2 Exercise 2: Consider a Markov chain with transition matrix P = 0 3 / 5 3 / 5 1 / 5 1 0 0 0 1 0 0 0 0 1 / 2 1 / 2 0 Find the stationary distribution of the chain. 1

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Solution: Since we have: P 5 = 0 . 32 0 . 404 0 . 148 0 . 128 0 . 64 0 . 2 0 . 12 0 . 04 0 . 64 0 . 2 0 . 12 0 . 04 0 . 2 0 . 48 0 . 16 0 . 16 is a regular matrix. We can find the limiting distribution by solving π = 0 . 32 π 1 + 0 . 64 π 2 + 0 . 64 π 3 + 0 . 2 π 4 = π 1 0 . 404 π 1 + 0 . 2 π 2 + 0 . 2 π 3 + 0 . 48 π 4 = π 2 0 . 148 π 1 + 0 . 12 π 2 + 0 . 12 π 3 + 0 . 16 π 4 = π 3 0 . 128 π 1 + 0 . 04 π 2 + 0 . 04 π 3 + 0 . 16 π 4 = π 4 π 1 + π 2 + π 3 + π 4 = 1 So, the stationary distribution is π = ( 5 11 . 7 22 , 3 22 , 1 11 ) 3 Exercise 3: The number of customers arriving at a coffee shop can be modeled by a Poison process with intensity λ = 15 customers per hour. a/ Find the probability that there are 7 customers between 10:00 and 10:20. b/ Find the probability that there are 3 customers between 10:00 and 10:20 and 7 customers between 10:20 and 11:00. c/ Find the probability of no arrivals between 10:15 and 10:30. Solution: We have: λ = 15 Denoted that X is the number of arrivals in that interval. a/ The length of time, 10:20 - 10:00 = 20 minutes = 1 3 hour. Thus, the probability that there are 7 customers between 10:00 and 10:20 is: P ( X 1 / 3 = 7) = e - 15 × 1 / 3 × (15 × 1 3 ) 7 7! = 0 . 1044 b/ The length of time, 11:00 - 10:20 = 40 minutes = 2 3 hour. The probability that there are 3 customers between 10:00 and 10:20 and 7 customers between 10:20 and 11:00 is: P ( X 1 / 3 = 3 , X 2 / 3 = 7)
= P ( X 1 / 3 = 3) × P ( X 2 / 3 = 7) = e - 15 × 1 / 3 × (15 × 1 3 ) 3 3! × e - 15 × 2 / 3 × (15 × 2 3 ) 7 7! = 0 . 0126 c/ The length of time, 10:30 - 10:15 = 15 minutes = 1 4 hour. The probability of no arrivals between 10:15 and 10:30 is: P ( X 1 / 4 = 0) = e - 15 × 1 / 4 = 0 . 0235 4 Exercise 4: A machine is in one of four states (F, G, H, I) and migrates annually among them according to a Markov process with transition matrix: F G H I F 0.2 0.8 0 0 G 0.5 0 0.5 0 H 0.75 0 0 0.25 I 1 0 0 0 At time 0, the machine is in state F. A salvage company will pay \$500 at the end of 3 years only if the machine is in state F (the salvage company will not pay any money if the machine is in other states). Compute the expected money paid by the salvage company after 3 years.

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