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STAT 312 Spring 2024 Investigation 5: Random rendezvous (assigned on Wed May 8, due on Thur May 16) You may work with in a group of as many as three students on this assignment, provided that you all contribute to the work. Type your answers to these questions into a file. (You do not need to re-type the questions.) Then save that file as a pdf file, and submit the pdf file in Canvas. Only one member of the group should submit a pdf file, making sure that all group member names appear at the top of the document. Suppose two people (let’s call them Eponine and Cosette, in memory of my first two cats) agree to meet for lunch at a certain restaurant. But these are both important people who can’t be sure when they’ll be able to get away, so both arrival times are random variables. If each person agrees to wait exactly fifteen minutes for the other before giving up and leaving, what is the probability that the two of them actually meet? Uniform distributions First suppose that the arrival time for each person, in minutes after noon, is uniformly distributed between 0 and 60. Also assume that the two arrival times are independent of each other. We will first use simulation to investigate what happens when this random process is repeated for a large number of days. The following graph resulted from simulating 5000 repetitions of this random process. a) What do the green dots represent? What do the red dots represent? - The green dots represent the simulated arrival times for Eponine - The red dots represent the stimulated arrival times for Cosette. © 2024 Allan J. Rossman STAT 312 – Spring 2024 1

This particular simulation of 5000 repetitions resulted in 2207 for which Eponine and Cosette successfully met. b) Use the simulation results to approximate the probability that Eponine and Cosette successfully meet. - P(meeting) = 2207/5000 = 0.4414 Because both distributions are uniform and independent of each other, exact probabilities can be calculated by determining the area of the region of interest as a fraction of the total area of the 60×60 square. c) Draw (by hand is fine) a 60×60 square to represent the sample space of all possible pairs of arrival times. Label one of the axes for Eponine’s arrival time and the other for Cosette’s arrival time. Then shade in the region of the square corresponding to the event that they arrive within 15 minutes of each other and therefore successfully meet. d) Use geometry to determine the area of the region in which Eponine and Cosette successfully meet. Then divide by the total area of the square to determine the probability that they meet. Is the actual probability within the error bound of the approximate probability from your simulation analysis? - The shaded region is a square with side length = 30 minutes. The total area of the square is 60(60) = 3600 square minutes. Therefore, P(meeting) = (30)(30) / 3600 = 1/12 e) Now let m represent the number of minutes that both people agree to wait, where m can be any real number between 0 and 60. Use geometry to express the probability that they successfully © 2024 Allan J. Rossman STAT 312 – Spring 2024 2

meet as a function of m . Also produce (and submit) a graph of this probability as a function of m , and comment on its behavior. - P(meeting) = (2m+30)^2 /3600 - f) Determine how long each person would have to agree to wait in order for this probability (of successfully meeting the other person) to be at least 0.5. Then determine how long each person would have to agree to wait in order for this probability to be at least 0.9. P(meeting) = 0.5 - M = 0.5 - (2m+30)^2 / 3600 - (2(0.5)+30)^2 / 3600 - = 10.36 minutes P(meeting) = 0.9 - M = 0.9 - (2m+30)^2 / 3600 - (2(0.9)+30)^2 / 3600 - = 24.49 minutes Normal distributions (Wait until after we have studied this topic on Mon May 13 to work on this section.) Now suppose that each person’s arrival time, in minutes after noon, follows a normal distribution with mean 30 and standard deviation 10. Continue to assume that they arrive independently of each other and that they agree to wait for 15 minutes. © 2024 Allan J. Rossman STAT 312 – Spring 2024 3

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Again we will first use simulation to investigate what happens when this random process is repeated for a large number of days. The following graph resulted from simulating 5000 repetitions of this random process. g) Describe how this graph differs from the previous one, and explain why this makes sense. This particular simulation of 5000 repetitions resulted in 3553 for which Eponine and Cosette successfully met. - Because this follow a normal distribution, the dots are situated in a closer circle. When graphed, this would form a normal distribution with the mean as the peak and two standard deviations. h) Use the simulation results to approximate the probability that Eponine and Cosette successfully meet. Also provide an error bound for this approximate probability. - P(meeting) = 3553/5000 = 0.7106 The exact probability that they meet can be determined from a normal distribution, using the following theoretical result: • Suppose that X and Y have independent normal distributions with means μ x and μ y and standard deviations σ x and σ y . Then the difference (X – Y) also has a normal distribution. o E(X – Y) = μ x – μ y o Var(X – Y) = σ x 2 + σ y 2 i) Report the probability distribution of the difference (not absolute difference) in the arrival times of Eponine and Cosette. (Be sure to include the mean and variance of the difference.) - X = T(Eponine) = N (30, 102) - Y = T(Cosette) = N (30, 102) - So the difference is 0, because 30 – 30. o E(X. – Y) = 30 – 30 - Var(X – Y) = 102 + 102 = 304 - These are both independent of one another. - SO =,t eh final difference is N(0, 204). © 2024 Allan J. Rossman STAT 312 – Spring 2024 4

j) Use appropriate normal probability calculations to determine the probability that the two people successfully meet. Also report the values of the appropriate z -scores. [ Hint : First express the probability that they successfully meet in terms of the random variable D.] - P(meet successfully) = P(-15 < X < 15) - Z score for -15 = (-15 – 0) / sqrt 204 = -1.05 - Z score for +15 = (15 -0) / sqrt 204 = 1.05 - P(Z = -1.05) = 0.1423 - P(Z = 1.05) = 0.8577 - 0.8577 – 0.1423 = 0.7154 - P(meet successfully) = 0.7154 k) Now let m represent the number of minutes that both people agree to wait, where m can be any real number. Determine the value of m so the probability of meeting is 0.9. - P(-m < X < m) = 0.9 - P(X > m ) = (1-0.9) / 2 = 0.05 using symmetry - Z score = m – 0 / sqrt(200 ) = m / 14.42 - P(Z > m / 14.42 ) = 1.645 - M = 14.42 (1.645) - M = 23.7209 l) Now suppose that Eponine and Cosette can only afford to wait for 15 minutes, but they want to have at least a 90% chance of successfully meeting. Continue to assume that their arrival times follow independent normal distributions with mean 30 and the same SD as each other. Determine how small that SD needs to be in order to meet their criteria. (As always, show your work.) - P(-15 < x < 15) = 0.9 - P(x > 15) = (1-0.9) / 2 - = 0.05 - Z score for x = 15= (15 – 0) / s - = 15/s where s = standard deviation of X. - P(Z ? 15/s) = 0.05 - 15/s = 1.645 - S = 15/1.645 - S = 9.12 minutes - SD = s / sqrt(2) - = 9.12 / sqrt(2) - 6.45 minutes © 2024 Allan J. Rossman STAT 312 – Spring 2024 5

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