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TEST AMS310.03 1--A FORM Fall 2003 1. PRINT YOUR NAME HERE______________________________ USE ONLY UPPER CASE LETTERS. UNDERLINE YOUR LAST NAME TWICE. 2. STUDENT ID NUMBER 3. CHECK TO MAKE SURE THAT YOUR TEST HAS 7 PAGES INCLUDING THIS ONE. 4. SHOW YOUR WORK FOR ALL QUESTIONS IN SECTIONS II. NO CREDIT WILL BE GIVEN FOR A NUMERICAL ANSWER WITHOUT AN ARGUMENT. POINTS 30 20 20 20 POINTS OFF POINTS EARNED SECTION I QUESTION II-1 QUESTION II-2 QUESTION II-3 QUESTION II-4 10 ----------------------------------------------------------------------------------------------TOTAL 100 - __________ = ___________(SCORE) 1 AMS310 TEST 1 Form A PAGE 2 OF 7 PRINT YOUR NAME HERE___________________________ Part I: Short answer questions. Very few calculations are required. Write the correct answer in the space provided by the question number. _______1. If there are 12 items in a box, how many samples of n=2 can we choose without replacement? Ans)) C2 12! 66 2!10! 12 _______2. P(A) = 0.2 , P(B) = 0.5, P(A B)= 0.05. P(A B) =____. Ans)) P( A B) P( A) P( B) P( A B) = 0.2+0.5-0.05 = 0.65 ________3. Suppose we toss a single die and the result is the number of dots observed. Then S={1,2,3,4,5,6}. Let A= {2,4,6}, and let B={3 ,4,5,6}. Then A B = __________. Ans)) B {1, 2} so, A B {1, 2, 4, 6} _______4. P(A)=0.5 P(B)= 0.6, P(A B) = 0.35 P(B|A)= Ans)) P( B | A) P( B A) P( A) 0.35 0.5 0.7 _______5. Given S= {1,2,3,4,5,6} the outcomes of the toss of 1 die. Let A={4,5,6} and let B={1,2,5}. Which statement below is correct? (a) A and B independent events. A and B are not mutually exclusive events. (b) A and B are dependent events. A and B are not mutually exclusive event. (c) A and B are independent. A and B are mutually exclusive. (d) A and B are dependent events. A and B are mutually exclusive. Ans)) (B) i) A B {5} therefore, A and B are not mutually exclusive. 1 1 1 ii) P( A) , P( B) and P( A B) P({5}) 2 2 6 1 1 1 1 P( A B) Therfore, A and B are dependent events then, P( A) P( B) 2 2 4 6 _______6. P(A|B) =0.1 P(A)=0.2 P(B)=0.6 P(A B)= ______ Ans)) P( A B) P( A | B) P( B) 0.1 0.6 0.06 2 AMS310 TEST 1 form A PAGE 3 OF 7 PRINT YOUR NAME HERE______________________ WRITE CORRECT ANSWER IN THE SPACE PROVIDED BY THE QUESTION NUMBER ________ 7. Given the values for F(x), the cumulative distribution function of X for values of X for which f(x)>0. Find P(X x F(x) Ans)) P( X 0 0.1 1.0). 0.5 0.4 1.0 0.5 1.5 0.6 2.0 0.8 3.0 1.0 1.0) 1 P( X 1.0) 1 P( X 0.5) 1 0.4 0.6 ________8. Evaluate x 0 ( x 5) 2 5x e x! 5 Ans)) x x 5x 5 2 ( x 5) e (x ) e x! x! 0 all : Variation formula of Poisson distribution with 2 5 then 5 = 2 5 __________ 9. X has Poisson distribution with =2. Use Table 2 to find Pr({X >3}). Ans)) P({X 3}) 1 P ({X 3}) 1 0.857 0.143 ___________ 10. Given the values of X and f(x) below . Find the value of , the mean of X. X f(x) -3 -2 0 0.2 0.2 0.2 4 0.4 Ans)) by definition of , x f ( x ) ( 3) (0.2) ( 2) (0.2) 0 (0.2) 4 (0.4) all 0.6 0.4 0 1.6 0.6 3 AMS310 TEST 1- PAGE 4 OF 7 PRINT YOUR NAME HERE______________________ LONGER QUESTIONS. FOR THESE QUESTIONS WHEREVER THERE IS A NUMERICAL ANSWER YOU NEED TO SHOW HOW YOU ARRIVED AT IT. YOUR SCORE WILL BE DETERMINED BY YOUR PRESENTATION OF YOUR METHOD. II- 1 . The length of "1 inch" nails, X, is a random variable with mean, = 1.0 inch. and standard deviation, =0.05 inches. (a) Use Chebyshev's inequality to determine (the upper limit to) the probability that the length of the nail is off by more than 0.25 inch. Hint: 0.25= k . Ans)) = 1.0 , =0.05 since 0.25= k = k 0.05 then k=5 Therefore, with P( X k ) 1 , P( X 1 k2 0.25) 1 52 1 25 0.04 (b) Use Chebyshev's inequality to determine the (lower limit to the) probability that a nail is between measures between 0.9 inches. and 1.1 inches. Hint: We want the probability that the nail length is on the interval from 0.9 inches to 1.1 inches. Ans)) P(0.9 X 1.1) 0.9 1 X 1 1.1 1 X 1 ) = P( 2 2) 0.05 0.05 0.05 0.05 X 1 P( 2) P ( X 1 2 0.05) 0.05 1 3 = 1 P( X 1 2 0.05) 1 2 0.75 2 4 P( (c) Find the range over which we expect to at least 91% of the nail lengths. (i) Probability = 0.91 = __________ (1- (1/k2) or (1/k2) 1 Ans)) = 1 2 k 1 1 9 1 3 (ii) k= ______ Ans)) since 1 2 therefore, 2 -> then k k k 100 k 10 (iii) k = ______ Ans)) k 10 3 10 5 3 100 1 6 (iv) Interval is from _______ to _________. 1 91 ) Ans)) P( X 1 therefore, the range is as follows. 6 100 1 1 1 5 7 X 1 X 1 X -> 1 -> 6 6 6 6 6 4 AMS310 TEST 1- PAGE 5 OF 7 PRINT YOUR NAME HERE______________________ LONGER QUESTIONS. FOR THESE QUESTIONS WHEREVER THERE IS A NUMERICAL ANSWER YOU NEED TO SHOW HOW YOU ARRIVED AT IT. YOUR SCORE WILL BE DETERMINED BY YOUR PRESENTATION OF YOUR METHOD. II-2. Two tests are done before a product leaves the factory. In the past the probability of failing the first test equals 0.01 and the probability of failing the second test equals 0.02 and the probability of failing both tests equals 0.005. Let F1 denote the event {an item fails the first test} and let F2 denote the event { an item fails the second test} (a) What is the probability that an item fails the second test given it has failed the first test? Ans)) P( F1) 0.01 , P( F 2) 0.02 , P( F1 F 2) 0.005 P( F1 F 2) 0.005 P( F 2 | F1) 0.5 P( F1) 0.01 (b) Show that an item failing the second test {F2} is dependent on the item failing the first test {F1}. Ans)) i) P( F 2 | F1) 0.5 P( F 2) 0.02 ii) P( F1) P( F 2) 0.01 0.02 0.0002 0.005 P( F1 F 2) (c) What is the probability that an item fails at least one test? Ans)) P( F1 F 2) P( F1) P( F 2) P( F1 F 2) = 0.01 0.02 0.005 0.025 (d)What is the probability an item passes both tests? Ans)) P(F1 F 2) P( F1 F 2) 1 P( F1 F 2) 1 0.025 0.975 (e) What is the probability of passing at least one test?, i.e., P( F1 (Hint: passing at least one test is the opposite of failing both tests) Ans)) P( F1 F 2) P ( F1 F 2) 1 P( F1 F 2 }? F 2) 1 0.005 0.995 5 AMS310 TEST 1- PAGE 6 OF 7 PRINT YOUR NAME HERE_____________________ LONGER QUESTIONS: FOR THESE QUESTIONS WHEREVER THERE IS A NUMERICAL ANSWER YOU NEED TO SHOW HOW YOU ARRIVED AT IT. YOUR SCORE WILL BE DETERMINED BY YOUR PRESENTATION OF YOUR METHOD. II-3. We are told that 5% of the computers sold will be returned for some repairs within a year of their sale. In the past day 20 computers are sold. Thus the distribution of X, the number computers of in a sample of 20 needing repairs within a year has a binomial distribution. (a)(i) What is the mean number of computers in samples of 20 that will need repairs within a year? Ans)) n 20 , p 0.05 , np 20 0.05 1.0 (a)(ii)What is the standard deviation of X? Ans)) 2 np(1 p) 20 0.05 0.95 0.95 0.95 0.97 Therefore, (a)(iii) Let Y denote the number of computers in a sample of 20 which will not be returned within a year. What is the standard deviation of Y? Ans)) P(Y ) 1 P( X ) 2 y npy (1 py ) 20 0.95 0.05 0.95 y 0.95 0.97 (b)What is the probability that more than 2 of these 20 computers will need repairs within a year ? Ans)) P( X 2) 1 P( X 2) 1 B(2;20,0.05) 1 0.9245 0.0755 (c) What is the probability that less than 18 out of 20 computers will NOT need repairs within a year? Ans)) P(Y 18) P(Y 17) B(17;20,0.95) 0.0755 Or P(Y 18) P(Y 17) P( X 3) P( X 2) 0.0755 since x y 20 (d) Suppose that in fact 3 out of this sample of 20 computers are returned within a year. Use your result in part (b) to comment on whether having 3 or more (or 10%) of one's sample of 20 needing repair is an unusual event. (Unusual is an event with probability less than 0.05). Write at least one sentence. Ans)) P( X 3) P( X 2) 0.0755 : not less than 0.05 (5%) Therefore, this is not an unusual event since the probability of 3 or more out of 20 needing repair equals 0.075 > 0.05 6 AMS310 TEST 1 PAGE 7 OF 7 II-4 Prove Bayes Theorem for k=2. P(A |B) = PRINT YOUR NAME HERE_______________________ P( B | A) P( A) P( B | A) P( A) P( B | A ) P( A ) P( A B) P( A B) P( A ' B) P ( B | A) P ( A) P( B | A) P( A) P ( B | A) P ( A) Ans)) P( A | B) (i) (ii) (iii) (iv) P( A B) P( B) P(A B) = P(B|A)P(A) (definition of P(B|A)). Similarly P(A' B) = P(B|A')P(A') (definition of P(B|A'). Since P(B) = P(A B) + P(A' B) , we have from (I) and (ii) that P(B)= P(B|A)P(A) + P(B|A') P(A') By definition of conditional probability P(A|B)= P(A B) / P(B). Substitute (i) for P(A B) and (iii) for P(B) and we obtain the result given by Bayes' Theorem. 7 AMS310.03 AMS310 TEST 1 Form A PAGE 2 OF 7 TEST 1-- makeup Fall 2003 PRINT YOUR NAME HERE___________________________ Part I: Short answer questions. Very few calculations are required. Write the correct answer in the space provided by the question number. _______1. If there are 12 items in a box, how many samples of n=2 can we choose with replacement? Ans)) 12 12 144 _______2. P(A) = 0.2 , P(B) = 0.6, P(A B)= 0.1 P(A B) =____. Ans)) P( A B) P( A) P( B) P( A B) = 0.2+0.6-0.1 = 0.7 ________3. Suppose we toss a single die and the result is the number of dots observed. Then S={1,2,3,4,5,6}. Let A= {2,4,6}, and let B={3 ,4,5,6}. Then A B = __________. Ans)) A {1,3,5} , B {1, 2} so, A B {1, 2,3,5} _______4. P(A)=0.5 P(B)= 0.6, P(A B) = 0.35 P(B|A)= Ans)) P( B | A) P( B A) P( A) 0.35 0.5 0.7 _______5. Given S= {1,2,3,4,5,6} the outcomes of the toss of 1 die. Let A={4,5,6} and let B={1,2}. Which statement below is correct? (a) A and B independent events. A and B are not mutually exclusive events. (b) A and B are dependent events. A and B are not mutually exclusive event. (c) A and B are independent. A and B are mutually exclusive. (d) A and B are dependent events. A and B are mutually exclusive. Ans)) (D) i) A therefore, A and B are mutually exclusive. 1 1 ii) P( A) , P( B) and P( A B) P( ) 0 2 3 1 1 1 0 P( A B) Therfore, A and B are dependent events then, P( A) P( B) 2 3 6 B _______6. P(A|B) =0.1 P(A)=0.2 P(B)=0.6 P(A B)= ______ Ans)) P( A B) P( A | B) P( B) 0.1 0.6 0.06 8 AMS310 TEST 1 form A PAGE 3 OF 7 PRINT YOUR NAME HERE______________________ WRITE CORRECT ANSWER IN THE SPACE PROVIDED BY THE QUESTION NUMBER ________ 7. Given the values for F(x), the cumulative distribution function of X for values of X for which f(x)>0. Find P(X x F(x) Ans)) P( X 0 0.1 1.2). 0.5 0.4 1.0 0.5 1.5 0.6 2.0 0.8 3.0 1.0 1.2) 1 P( X 1.0) 1 0.5 0.5 ________8. Evaluate x 0 x 5x e x! 5 Ans)) x x 5x 5 x e x e x! x! 0 all : Mean formula of Poisson distribution with 5 then 5 = 2 5 __________ 9. X has Poisson distribution with =2. Use Table 2 to find Pr({1<X<4}). Ans)) P({1 X 4}) P (2 X 3) P ( X 3) P (P 2) P( X 3) P( X 1) 0.857 0.406 0.451 ___________ 10. Given the values of X and f(x) below . The value of =1, Calculate the value of X f(x) -3.5 -1.5 0 0.2 0.2 0.2 2 2 . 5 0.4 Ans)) by definition of 2 all . (x )2 f ( x ) 2 2 ( 3.5 1)2 (0.2) ( 1.5 1)2 (0.2) (0 1) (0.2) (5 1)(0.4) (20.25) 0.2 (6.25) 0.2 1 0.2 16 0.4 11.9 9 AMS310 TEST 1- PAGE 6 OF 7 PRINT YOUR NAME HERE_____________________ LONGER QUESTIONS: FOR THESE QUESTIONS WHEREVER THERE IS A NUMERICAL ANSWER YOU NEED TO SHOW HOW YOU ARRIVED AT IT. YOUR SCORE WILL BE DETERMINED BY YOUR PRESENTATION OF YOUR METHOD. II-3. We are told that 10% of the computers sold will be returned for some repairs within a year of their sale. In the past day 20 computers are sold. Thus the distribution of X, the number of computers in a sample of 20 needing repairs within a year has a binomial distribution. (a)(i) What is the mean number of computers in samples of 20 that will need repairs within a year? Ans)) n 20 , p 0.1 , np 20 0.1 2.0 (a)(ii)What is the standard deviation of X? Ans)) 2 np(1 p) 20 0.1 0.9 1.8 1.8 1.34 Therefore, (a)(iii) Let Y denote the number of computers in a sample of 20 which will not be returned within a year. What is the standard deviation of Y? Ans)) P(Y ) 1 P( X ) 2 y npy (1 py ) 20 0.9 0.1 1.8 y 1.8 1.34 (b)What is the probability that more than 2 of these 20 computers will need repairs within a year ? Ans)) P( X 2) 1 P( X 2) 1 B(2;20,0.1) 1 0.6769=0.3231 (c) What is the probability that less than 19 out of 20 computers do NOT need repairs within a year? Ans)) P(Y 19) P(Y 18) B(18;20,0.9) 0.6083 (d) Suppose that in fact 1 out of this sample of 20 computers are returned within a year. Use your result in part (b) to comment on whether having 1 or more (or 5%) of one's sample of 20 needing repair is an unusual event. (Unusual is an event with probability less than 0.05). Write at least one sentence. Ans)) P( X 1) P(Y 19) 0.6083 : not less than 0.05 (5%) Therefore, this is not an unusual event since the probability of 1 or more out of 20 needing repair equals 0.075 > 0.05 10
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: AMS310 TEST 2 Fall 2003 FORM A 1. PRINT YOUR NAME HERE_ USE ONLY UPPER CASE LETTERS. UNDERLINE YOUR LAST NAME TWICE. 2. Write your student ID number here_ 3. CHECK TO MAKE SURE THAT YOUR TEST HAS 8 PAGES INCLUDING THIS ONE. 4. SHOW YOUR WORK FO...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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s06_prac_final_&_sol_eco2302
Path: UT Dallas >> ECON >> 2302 Spring, 2006
Description: MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Which of the following is the best definition of economics? A) The study of how consumers spend their income B) The study of how the federal gov...
s06_prac_exam2_eco2302
Path: UT Dallas >> ECON >> 2302 Spring, 2006
Description: Economics 2302 Principles of Microeconomics Professor Isaac McFarlin Practice Examination 2 University of Texas at Dallas Student Name (print) _ Student ID: xxx xx - _ Student Signature (required) __ INSTRUCTIONS: The exam is closed book. You may...
ps1_econ2302
Path: UT Dallas >> ECON >> 2302 Spring, 2006
Description: Economics 2302 Professor McFarlin Problem Set 1 Principles of Microeconomics University of Texas at Dallas [Note: Problem sets in this class will not be collected or graded.] Economics is a quantitative social science. If you take a moment and scan...
Chapter 8 true false
Path: Mines >> EBGN >> 312 Fall, 2002
Description: Chapter 8: Aggregate Expenditure and Equilibrium Output True or False 1. Consumption is a highly volatile component of aggregate expenditure. TRUE FALSE 2. The marginal propensity to consume is the slope of the aggregate expenditure function (C +...
Chapter 9 hw4
Path: Mines >> EBGN >> 312 Fall, 2002
Description: Chapter 9: The Government and Fiscal Policy Multiple Choice 1. In macroeconomics, the debate over the role of government includes the views of Keynesian economists who believe that: The government is incapable of stabilizing the economy. The govern...
externality and tax graphs
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: ...
Home work Statistics 220
Path: Carnegie Mellon >> STAT >> 220 Fall, 2005
Description: Homework Assignment Statistics 220, Fall 2005 Homework Assignment #1 Problems from Devore: 1. 2. 3. 4. 5. 6. 7. Problem 1.10 Problem 1.36 Problem 1.38 Problem 1.41 Problem 1.42 Problem 1.44 Problem 1.51 Homework Assignment #2 Due September 14 Show y...
PrevExam-1 part4
Path: Mines >> EBGN >> 312 Fall, 2002
Description: ...
ps4_ans_econ2302
Path: UT Dallas >> ECON >> 2302 Spring, 2006
Description: Economics 2302 Professor McFarlin Solution to problem Set 4 Principles of Microeconomics University of Texas at Dallas 1. Here\'s a table with all the info: L 0 1 2 3 4 5 6 7 8 Q 0 30 70 100 120 130 135 140 140 MPL 30 40 30 20 10 5 5 0 APL 30 35 33...
ps2_econ2302
Path: UT Dallas >> ECON >> 2302 Spring, 2006
Description: Economics 2302 Professor McFarlin Problem Set 2 Principles of Microeconomics University of Texas at Dallas [Note: Problem sets in this class will not be collected or graded.] 1. Consider an economy comprised of Jack and Jill, each of whom is capab...
PrevExam-1
Path: Mines >> EBGN >> 312 Fall, 2002
Description: ...
CPSC 303 Syllabus Numerical Approximation and Discretization 2007 Class Home Page
Path: UBC >> CPSC >> 303 Winter, 2007
Description: CPSC 303: Numerical Approximation and Discretization 2007 Class Home Page CPSC 303, Term 2, Winter 2007-2008 Class Home Page Numerical Approximation and Discretization Contents of this page Late breaking news Things to print r Handouts r H...
1.07_hw.0
Path: UBC >> CPSC >> 303 Winter, 2007
Description: Numerical Approximation and Discretization CPSC 303, Term 2, Winter 20072008 Assigned Monday January 7. Not due for grading. Submit by Thursday, January 24 if you wish to have the TA check it. Homework #0 This homework is designed to give you a litt...
1.07 soln.hw1
Path: UBC >> CPSC >> 303 Winter, 2007
Description: Numerical Approximation and Discretization CPSC 303, Term 2, Winter 20072008 Homework #1 Solutions 1. Weather Simulation. (10 points) Imagine trying to predict the weather in Vancouver tomorrow using a computer simulation. What might be involved in...
1.21 Linear Algebra Fact Sheet
Path: UBC >> CPSC >> 303 Winter, 2007
Description: Numerical Approximation and Discretization gg QHQD erm PD inter PHHT{PHHU Linear Algebra Review Notation: intry ppers in row nd olumn of F hen tlking out mtriesD rememerX rows then olumnsF4 etor P R hs rows @ olumn vetorAF intry ppers in row F p...
1.27 Interpolation Error Examples
Path: UBC >> CPSC >> 303 Winter, 2007
Description: Numerical Approximation and Discretization CPSC 303, Term 2, Winter 20072008 Interpolation Error Examples Consider a problem similar to what we studied using interpolateSine.m: interpolate f (x) = sin(2x) on the interval x [0, +1]. Our error formul...
1.21_hw.2
Path: UBC >> CPSC >> 303 Winter, 2007
Description: Numerical Approximation and Discretization. CPSC 303, Term 2, Winter 20072008. Assigned Monday January 21. Due Monday February 4. Maximum 2 late days. Homework #2 1. Paper and Pencil. (10 points) (a) Imagine that we have already computed an interpol...
1.24 Approximation & Error Example
Path: UBC >> CPSC >> 303 Winter, 2007
Description: Numerical Approximation and Discretization CPSC 303, Term 2, Winter 20072008 Approximations & Errors Example There are two sources of error we need to take into account when we are designing and analyzing numerical algorithms: data error (which occu...
Answers to Ch 2 Problems
Path: JMU >> ECON >> 201 Fall, 2007
Description: Answers to Problems: Chapter 2 1. In time it takes Ted to wash a car he can wax one-third of a car. So his opportunity cost of washing one car is one-third of a wax job. In the time it takes Tom to wash a car, he can wax one-half of a car. So his opp...
Ch7 Answer to Problems
Path: JMU >> ECON >> 201 Fall, 2007
Description: ...
chapter 3 answers
Path: JMU >> ECON >> 201 Fall, 2007
Description: Answers to Problems 1a. Substitutes b. Complements c. Probably substitutes for most people, but complements for some others who like to eat ice cream and chocolate together. d. Substitutes. 2. The supply curve would shift: a. Right. The discovery is ...
Ch 11 Answers to Problems
Path: JMU >> ECON >> 201 Fall, 2007
Description: Answers to Problems 1a. All others Study a lot Study a lot -5 for Sam -5 for others Sam Study a little -6 for Sam 10 for others -1 for Sam -1 for others Study a little 10 for Sam -6 for others b. All study a lot and receive an average grade. From th...
Ch 10 Answers to Problems
Path: JMU >> ECON >> 201 Fall, 2007
Description: Answers to Problems 1. As shown in the following table, Volvo\'s greater production volume gives it substantially lower average production cost, and this advantage helps explain why Volvo\'s market share has in fact been growing relative to Saab\'s. Saa...
Ch 6 Answers to Problems
Path: JMU >> ECON >> 201 Fall, 2007
Description: Chapter 6 Answers to Problems 1. If the price of a fossil is below $5.40, Zoe should devote all her time to photography because when the price is, say, $5 per fossil, an hour spent looking for fossils will give her 5($5) = $25, or $2 less than she\'d ...
Syllabus
Path: UCSB >> ECON >> 01 Summer, 2007
Description: Econ 1: Principles of Microeconomics I. Administrative Information Lecturer: Francisco Azeredo Office: NH 2028 Teaching assistant: Ben Hansen Office: NH 2044 Class: MTW TA section: R R 9:30 10:55am 9:30 10:55am 12:30 1:55 pm Office Hours: TW 11:00...
Homework3
Path: UCSB >> ECON >> 01 Summer, 2007
Description: UNIVERSITY OF CALIFORNIA SANTA BARBARA Economics 1 Principles of Economics Summer 2007 Session A Homework #3 I. Problems: Chapter 5 I.1 Book Problems Francisco Azeredo Ben Hansen Please place your answers in the space below. Take the time to draw ...
