131A_1_ee131A_1_131A_1_prac-final-04-sol

131A_1_ee131A_1_131A_1_prac-final-04-sol - EE131A: Practice...

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EE131A: Practice Problems for Final Exam The exam will have 6 questions More than 6 problems have been provided for practice purposes Problem 1. ( 25 pts ) (a) (10 pts) Students taking a test have to answer 4 out of 5 questions, not in any specific order. The questions are numbered 1–5. Assuming that a student chooses questions randomly, what is the probability that she will answer questions 2–5 (again, not in any specific order)? (b) (15 pts) Now suppose that the questions are multiple choice, and each question has 4 possible answers, out of which exactly one choice is the correct answer. For each correct answer, students get 10 points, but for each incorrect answer they lose 2 points. One of the students has not studied at all and he is just guessing his answers. What is the probability that he will score at least 20 points on the 4 questions he answers? Sol. (a) P(problem 2,3,4,5 is choosen)= 1 0 @ 5 4 1 A = 1 5 (b) If the students gives 0 correct answer, he gets ( - 2)(4) = - 8 points; else if he gives 1 correct answer, he gets 10 + ( - 2)(3) = 4 points; else if he gives 2 correct answer, he gets 20 - 4 = 16 points; else if he gives 3 correct answer, he gets 30 - 2 = 28 points. .. So, P(he scores at least 20 points) = P(he gives at least 3 correct answer) = p 4 3 P p 3 (1 - p ) + p 4 = 4(0 . 25 3 )(0 . 75) + 0 . 25 4 = 0 . 05 Problem 2. (25 pts) (a) (15 pts) A continuous random variable X that takes on only positive values (i.e., P [ X 0] = 0 ) has cdf F X ( x ) and pdf f X ( x ) . Find the cdf and pdf of the random variable Y = ln X in terms of F X ( x ) and f X ( x ) . (b) (10 pts) In a large hotel it is known that each guest has a 1% chance of losing his key. If a key is lost, the guest must pay a fine of $20. A group of 200 UCLA students checks into the hotel for a meeting. Using the Poisson approximation , compute the probability that the hotel will collect at least $60 in lost key fines from the group. Sol. (a) F Y ( y ) = P ( Y y ) = P (ln X y ) = P ( X e y ) = F X ( e y ) f Y ( y ) = dF Y ( y ) dy = F p X ( e y ) d dy e y = f X ( e y ) e y (b) Let X be the the number lost keys by all the 200 students. Then α = E [ X ] = 200 × 0 . 01 = 2 . For the hotel to collect at least $60, at least three students must lose their keys, or X 3 . Hence, P ( X 3) = e - 2 s k =3 2 k k ! = 1 - e - 2 2 s k =0 2 k k ! = 1 - 5 × e - 2 = 0 . 32332 Note that if no approximation is made, then we have to use Binomial law, and the calculations will go as follows: P ( at least $60 in fines ) = P ( X 3) = 1 - 2 s i =0 p 200 i P (0 . 01) i (0 . 99) 200 - i , 1
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which equals (accurate up to 5 decimal places) 0 . 32332 . Hence, the Poisson approximation is excellent for this situation. Problem 3.
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131A_1_ee131A_1_131A_1_prac-final-04-sol - EE131A: Practice...

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