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Unformatted text preview: CHEN 3010 Applied Data Analysis Fall 2009 Name:________________________________ Midterm Examination 1 Solutions Note: This is the solution for version A (except for problem 7). Other versions were just mixed up questions and numbers changed slightly. 1) (10 pts) Although not endorsed by any of the ChBE faculty, hitchhiking can be an efficient way to get around without a car. You are hitchhiking your way back to mom and pop’s place and cars are few and far between. At this time of day and in this particular spot there are approximately 6 cars that pass by every 15 minutes. a. (5 pts) If you were to just stand there and observe for a while (no thumb stuck out), what is the probability that in a 10minute time period there would be more than 5 cars that passed by? Solution: 6 0.4 / where
3 2 1
. Poisson distribution:
5 1 5 4 ! 10 0 4.0 1 5 1 0.018 0 1 0.073 2 0.147 3 0.195 4 0.195 5 0.156 . b. (5 pts) Now you stick your thumb out. Assuming that every car that passes by will pick you up (a very bad assumption), what is the probability that you will have to wait longer than 5 minutes to get a ride? Solution: The cumulative exponential distribution is 1 The probability that you will be picked up in less than 5 minutes is: 5 1
. ·5 0.865; therefore, the probability that you wait longer than 5 minutes would be 0.135. 2) (10 pts) The engines of an airplane operate independently. The probability that an individual engine will operate properly for a given trip is 0.95. A plane will be able to complete a trip successfully if AT LEAST half of its engines operate for the entire trip. Determine whether a fourengine or a twoengine plane has the higher probability of a successful trip. State the respective probabilities for full credit. Solution: CHEN 3010 Midterm Exam 1 Fall 2009 Name:________________________________ A 2engine plane will work if one or two of its engines are working. described by the binomial distribution: 1 1 2 0.95 0.05 1 1 2 2 0.95 0.05 2 1 2 0.9975 This can be Similarly, for a 4engine plane we need 2, 3, or 4 engines to be working properly: 1 4 0.95 0.05 2 4 0.95 0.05 3 4 0.95 0.05 4 0.9995 The 4engine plane therefore has a greater probability of success. 3) (10 pts) Imagine that you are able to successfully ride The Wave at Water World on 70% of your tries. Now suppose that you want to get at least 4 successful Wave rides during your trip to Water World. How many times will you have to attempt The Wave in order to be at least 75% sure of meeting your goal? Solution: This is a negative binomial problem because you need 4 successes. We wish to determine the number of trials (attempts) that we will have to plan for to be at least 75% sure of getting 4 successes. The following equation describes the negative binomial distribution, where r=4 successes and f(x) is the probability that you will have x number of attempts in order to get those 4 successes: 1 1 1 Thus, the probability that the 4th success will come on the 4th ride (x=4) is: 0.240. The probability that the 4th success will come on the 5th ride is: 0.288 (cumulative of 0.528). The probability that the 4th success will come on the 6th ride is: 0.216 (cumulative of 0.74). The probability that the 4th success will come on the 7th ride is: 0.130 (cumulative of 0.874). Therefore, to be 75% sure, you should ride The Wave 7 times. CHEN 3010 Midterm Exam 1 Fall 2009 Name:________________________________ 4) (15 pts) A review panel is being assembled to review grant proposals in the field of biomedical engineering. The panel will have exactly 6 members and there must be at least 2 engineers and at least 2 biologists. The organizer can choose from 8 engineers and 12 biologists. a. (6 pts) How many different suitable (at least 2 engineers and at least 2 biologists) can be assembled? Solution: The only combinations of panels that can be assembled are panels that contain 2 engineers and 4 biologists, 3 engineers and 3 biologists, or 4 engineers and 2 biologists. There are 8C2x12C4 = 13860 ways to make a panel containing 2 engineers and 4 biologists. There are 8C3x12C3 = 12320 ways to make a team containing 3 engineers and 3 biologists. There are 8C4x12C2 = 4620 ways to make a team containing 4 engineers and 2 biologists. This is a total of 13860 + 12320 + 4620 = 30800 ways to form panels with these restrictions. b. (9 pts) If Dr. Pasteur is one of the 12 biologists that can be selected, what is the probability that Dr. Pasteur will be selected for the panel, assuming that the final panel is randomly selected (but still meets the requirement of having at least 2 engineers and at least 2 biologists)? We want to calculate the total number of possible panels that would have Dr. Pasteur as one of the biologists. This means that the combinations of panels that can be assembled are panels that contain 2 engineers, 3 biologists, and Dr. Pasteur; 3 engineers, 2 biologists, and Dr. Pasteur; and 4 engineers, 1 biologist, and Dr. Pasteur. There are 8C2x11C3x1 = 4620 ways to make a panel with 2 engineers, 3 biologists, and Dr. Pasteur. There are 8C3x11C2x1 = 3080 ways to make a panel with 3 engineers, 2 biologists, and Dr. Pasteur. There are 8C4x11C1x1 = 770 ways to make a panel with 4 engineers, 1 biologist, and Dr. Pasteur. Thus, there are 4620 + 3080 + 770 = 8470 possible panels with these restraints. The probability that Dr. Pasteur is on the panel is thus 8470/30800 = 0.275. CHEN 3010 Midterm Exam 1 Fall 2009 Name:________________________________ 5) (20 pts) The concentration of a reactant is a random variable with probability density function: 0 1 0 a. (4 pts) Find k such that is a probability density function. Solution: To be a probability density function, the area under the curve must be equal to 1: 1 2 1 3 1 b. (4 pts) What is the probability that the concentration is greater than 0.7? Solution: 0.7 6 6
. . . . . . . c. (12 pts) Find the probability that the concentration is within 0.5 standard deviations (±0.5σ) of the mean. Solution: To solve this, we must first determine the mean and standard deviation:
. . 6 0.5 6 20 0.5 . 0.224 0.5 3 6 4
. .
. 4 5 0.5 0.112 Now that mean and standard deviation are known, we can determine the probability that the concentration is within 0.5 standard deviations (±0.5σ) of the mean:
. . . 0.388 0.612 6
. 6 2 . 3 .
. CHEN 3010 Midterm Exam 1 Fall 2009 Name:________________________________ 6) (15 pts) A 500page book contains 250 sheets of paper. The thickness of the paper used to manufacture the book is a normal random variable and has mean 0.08 mm and standard deviation 0.01 mm. a. (7 pts) What proportion of the pages are between 0.06 and 0.07 mm? Solution: We can convert 0.06 mm and 0.07 mm to standard normal variables (zvalues): 0.06 0.08 0.01 2.0 0.07 0.08 0.01 1.0 . . The cumulative frequencies can be looked up in the z tables in the back of your book: 0.06 0.07 0.159 0.023 . b. (8 pts) What is the probability that a randomly chosen book is more than 20.2 mm thick (not including the covers)? Solution: This is a problem related to functions of random variables. represent the total thickness of of a book of 250 sheets. 250 · 250 · Therefore: 20.2 20.2 20 2.5 0.08 1 0.08 . 6.25 20 2.5 Let Y 7) (20 pts) Determine the distribution of the data on the following page. You may use any method you like but you must show full work (plugging into your calculator is NOT showing your work). Show work on the attached sheet next to the data or attach your own sheet to the exam. Solution: See attached Problem 7 Solution. Determine th di t ib ti t i the distribution of th f these d t using any method data, i th d without without using a computer or calculator. Version A 87 99 82 59 97 91 70 57 74 66 77 51 74 68 95 58 61 54 71 50 60 76 86 99 84 84 53 68 95 85 76 78 65 73 94 91 65 84 68 88 Version B 81 97 79 66 91 84 73 64 74 75 76 56 74 76 88 65 67 61 73 51 67 76 81 94 80 80 59 72 87 80 76 76 75 74 87 84 70 80 82 82 Version C 86 84 92 88 82 90 96 85 89 78 85 78 91 84 79 90 90 90 93 82 76 61 91 96 85 100 93 87 96 91 86 83 57 98 79 93 99 75 84 97 50 60 70 80 90 100 Version C Bin Frequency 50 0 60 1 70 1 80 6 90 18 100 14 More 0 Version B Bin Frequency 50 0 60 3 70 7 80 17 90 10 100 3 More 0 Bin Frequency 50 0 60 8 70 7 80 9 90 8 100 8 More 0 Version A
10 Version B V i Version C 18 20 9 16 18 8
14 16 7
12 14 6 Frequency equency ency
Frequency equency ncy 10 Frequency uency y 12 10 5 8 8 4
6 6 3
4 4 2
2 1 2 0 0 50 50 60 70 80 Bin 90 100 More 60 70 80 Bin 90 100 More 0 50 60 70 80 Bin 90 100 More *Uniform Distribution *Normal Distribution *Gamma or Weibull Distribution ...
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This note was uploaded on 04/25/2010 for the course CHEN 3010 at Colorado.
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