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HOMEWORK 4 1. Ross Chapter 2, problems 2, 5, and 8 *Problem 2.2 (p. 33) 2. Consider an experiment that consists of six horses, numbered 1 through 6. running race. and suppose that the sample space is given by S = ( all orderings of (1,2,3,4,5,6)] Let A denote heevent that the number horse isamong the top three finishers, let Bdenotetheeventthat henumber 2horse ecmesin second, and let 'denote be event that the number horse comes in third (a) Describe the event A u B. How many outcomes are contained in this event? (b) How many outcomes are contained in the event AB? (b) How many outcomes are contained in the event ABC? (c) How many outcomes are contained in the event A u BC? *Problem 2.5 (p. 34) 5. The random variable X takes on one of the values 1,2,3, 4 with probabilities P[X = i) = ic, i = 1, 2,3,4 for some value c. Find P(2 X = 3). *Problem 2.8 (p. 34) 8. Find the expected value of the random variable specified in Exercise 5. 2. Ross Chapter 3, problems 3 and 8 *Problem 3.3 (p. 44) In Exercises 3-9 use simulation to approximate the following integrals. Compare your estimate with the exact answer if known. 3. So exp(e*)dx 4. 6'11-x2)312x 5. 6. 6xx(1+x)-2dx 7. Love-x2x 8. 9. [Hint: Let ly(x) = and use this function to equate the integral 0 if y>x to one in which both terms go from 0 to co.] 10. Use simulation to approximate Cov(U, e)). where U is uniform on (0, 1). Compare your approximation with the exact answer. *Problem 3.8 (p. 44) In Exercises 3-9 use simulation to approximate the following integrals. Compare your estimate with the exact answer if known. 3. So exp{e*)dx 4. 6'(11-x2)312x 5. S2ext2-dx 6. 6xx(1+x)-2dx 7. Love-2dx 8. 9. 60.6e-itr/dydx [Hint: Let ly(x) = and use this function to equate the integral to one in which both terms go from 0 to oo.] 10. Use simulation to approximate Cov(U, e).). where U is uniform on (0, 1). Compare your approximation with the exact answer. 3. Given parameter Lambda and a series of random numbers drawn from U(0,1), generate interarrival times using the exponential randomvariable. Use the built-in scientific calculator in your operating system. Try the following: Lambda and U=0.5 Lambda 4 and U 0.01 Lambda and U 0.2 Lambda and U 0.368 Question Completion Status: QUESTION 1 1. Arandom number generator uses the following algorithm: Xn- 3) mod 70 (a) If the seed value is 11, what are the next five values generated by the algorithm? ) Just based on the values you produced, would you consider this to be a good random number generator? Why or why not? What : me important attributes of a good random number generator for use in simulations? QUESTION N 2 A function of x is shown in the picture bekow. Suppore that a =3 and b 6. Using onsly the following realizations of - rough estimate of the definite intego zal of the function using the Monte method. un 2107 0.06, 3 2 1 QUESTION 3 3. Suppose you could construct a fair five-sided die, with sides labeled 1 through 5, and roll this die for a very large number of times. (a) What are the expectation and the variance of this experiment? (b) Write two lines of R code to simulate the results of 1000 rolls of the die. QUESTION 4 4. Acontinuous random variable has the following probability density function, f(x): f(x) = 2x in the interval o (a) Write an expression for the cumulative distribution function (b) What is the probability that a realization of X will be less than 0.4 ? - QUESTION 5 5. Customers arrive at a bank according to a homogeneous Poisson process at rate Lambda - 5 per hour. The bank opens at 9am, and TMA customers are waiting outside at the opening The first customer arrives at 9.1 hours (in other words, 9:06am). Given the following random numbers drawn from U(0,1), compute the inter-arrival times and arrival times of the reet four customers. Round to two decimal places (decimal hours, not minutes). Using the results, assume that there is one teller and that the service time for each customer is precisely 02 hours. Fill out the remainder of the table below. Values of U to use for customers 2 through 5: U2 = 0.78; Inter-arrival time U3 - 0.37; Inter-arrival time U4 0.47; Inter-arrival time U5 = 0.11; Inter-arrival time Customer # 1 2 3 4 5 Arrival time 9.1 Starts Service 9.1 Departs 9.3 QUESTION de ta ant am - Poizzon with Lamblada(1) shown in the figuare beekow. 20 20 15 15 10 10 5 5 5pm 7am 9am 2 pm 7am 5pm 9 am 2 pm (a) What are the highest and lowest values of Lambda (t)? (b) In the thinning process, what value of Lambda should you use in the following equation to generate arrivalsin your simulation: Lambda_ratio=lambdalt)/ Lambda Lambda= (c) What is "Lambda_ratio" at 9 am? (d) You 've implemented thinning in your simulation. The simulation generates a possible arrival at 2pm. Your thinning algorithm compares "Lambda_ ratio" to a random number "U" drawn from U(0,1). Suppose U = 0.37. Should you accept or reject this arrival? QUESTION You conduct an experiment of flipping a fair coin T times. What is the probability of getting exactly 3 heads? T/T Paragraph I - Arial 3 (12pt) T' T. e is QUESTION 8 8. Arandom variable x takes on one of the values 1,2,3,4 with probabilities - IC, wherei = 1,2,3,4 for some value of C (a) Find {{<<=X<=3} (b) Find the expected value E[X] (c) Describe how would you generate this random variable from U(0,1)

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Q1
X0 = 11
Xn = (7 ∗ Xn−1 + 3) mode 70
X1 = (7 ∗ X0 + 3) mode 70 = (7 ∗ 11 + 3) mode 70 = 10
X2 = (7 ∗ X1 + 3) mode 70 = (7 ∗ 10 + 3) mode 70 = 73
X3 = (7 ∗ X2 + 3) mode 70 = (7 ∗ 73 + 3) mode 70 = 24
X4 = (7 ∗ X3 + 3) mode 70 = (7 ∗ 24 + 3) mode 70 = 31
X5 = (7 ∗ X3 + 3) mode 70 = (7 ∗ 31 + 3) mode 70 = 10
It cannot be considered as good random generator, as the produced values are not uniformly distributed. The mode value is not a prime number, hence there would be an unqual probability of generating numbers. Hence, next value is dependent on the previous value, hence it is not a good property of random number generators.
There are two properties of a good random number generator:
1. Equal probability of generating numbers.
2. There must be no dependency on previous value in order to generate next value.
Q2
Area under the curve can be estimated by multiplying distance between ‘b’ and ‘a’ and average height of the poles.
Monte Carlo gives, average height 1.99.
Estimated Area is: 1.99 * (3-1) = 3.98...

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