Homework 3 Solutions - STAT 225 Fall 2016 Homework...

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Applied Calculus
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Chapter 8 / Exercise 53
Applied Calculus
Berresford/Rockett
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Unformatted text preview: STAT 225 Fall 2016-- Homework 3-------SOLUTIONS 1. There are several scenarios described below. For each of them, do the following (note: R.V. means random variable) (1) Define the R.V.--- that means something like, “Let X be the number of people who…..” (2) Define the distribution and parameter(s) of the R.V. (3) Give the support of the R.V. (4) Write the probability statement related to the information being sought. Do not calculate the probability. a) The Ohio Bureau of Motor Vehicles states that 7 out of 8 people pass the test. A test-taker wants to find out the probability that he will pass the written test in fewer than 4 tries. (1) ℎ ℎ ℎ . 7 (2) ~ ( = ) 8 (3) ∈ {1,2, … } (4) ( < 4) b) LAIMO manufacturing company makes parts for the auto industry. Approximately 3% of the parts it makes are defective. We are interested in calculating the probability that the third defective is the 20th one sampled. (1) ℎ ℎ 3 (2) ~ ( = 3, = 0.03) (3) ∈ {3,4, … } (4) ( = 20) c) A BigMart store is going to hire 3 new cashiers. It has a pool of 18 applicants (10 male, 8 female) for a for these 3 cashier jobs. The BigMart manager is interested in the probability that none of the positions are filled by females. (1) ℎ ℎ ℎ 3 ℎ . ( = #). (2) ~ ( = 18, = 3, = 8) ~ ( = 18, = 3, = 10) (3) ∈ {0,1,2,3} ∈ {0,1,2,3} (4) ( = 0) ( = 3) d) A gardener is inspecting her the fall flowers in her garden. She notices, on average, 4 bugs on a flower. She randomly picks one flower from her garden. She wants to know the probability that the flower she picked has at least one bug on it. (1) ℎ . (2) ~ ( = 4) (3) ∈ {0,1, … } (4) ( ≥ 1) e) A student is taking a true/false test that consists of 15 questions. Based on past performance the student has approximately a 70% chance of getting any individual question correct. We are interested in the probability that the student gets at least 60% of the questions on the test correct. (note: at least 60% of the questions correct means that at least 0.6 * 15 = 9 so at least 9 questions correct) (1) ℎ ℎ 15 . (2) ~ ( = 15, = 0.7) (3) ∈ {0,1, … 15} (4) ( ≥ 9) f) A certain radio station’s phone lines are busy approximately 95% of the time when trying to call during a contest. We are interested in finding the probability that the 4th time you call is the 1st time you get through during a contest. (1) ℎ ℎ ℎ ℎℎ. (2) ~ ( = 0.05) (3) ∈ {1,2, … } (4) ( = 4) 2. Suppose that 1% of the Blu-ray discs produced by a company are defective. You buy one of the Blu-ray discs and check it see if it is defective. a) What is a “success” in this story? What is the probability of a success? “” − . ℎ 0.01. b) Define X in terms of the story. What values can X take? – ℎ − ℎ ℎ. 0 1. c) What are the distribution and parameters of X? ~ ( = 0.01) d) Instead of buying just one Blu-ray disc, you have purchased 25 discs as gifts for your friends and family. What is the probability that 2 of the discs are defective? What distribution and parameter(s) are you using? 25 ~ ( = 25, = .01), ( = 2) = ( ) (0. 01)2 ∗ (0. 99)23 = 0.0238 2 e) How many discs would you need to purchase in order to be at least 90% sure that you have at least one defective disc? ~ (, = .01) ( ≥ 1) = 0.9 1 − ( = 0) = 0.9 1 − ( ) (0. 01)0 (0. 99) = 0.9 0 ∗ log(0.99) = log(0.1) log(0.1) = = 229.105 log(0.99) : = 230 3. A representative from Purdue Musical Organizations (PMO) randomly surveys people around various parts of campus. He is interested in finding people who attended a PMO performance last semester. From previous surveys, he knows that the probability a person on campus did not attend a PMO performance last semester was 0.73. a) Let M be the number of people surveyed until the representative finds a person who did attend a PMO performance last semester. What is the distribution, parameter(s) and support of M? ~( = 0.27). The support is {1,2,3,4, … }. b) What are the average and standard deviation of L? 1 1 [] = = .27 = 3.7037 1− 1−.27 .73 () = √ 2 = √ .272 = √.0729 = 3.1644 c) What is the probability he surveys at least 5 people until he finds a person who attended a PMO performance last semester? There are two methods of calculating the desired probability. The first uses the so-called tail property for the Geometric distribution. The second method is more direct, but tedious way. Method 1: Tail Property ( ≥ 5) = ( > 4) = (1 − )4 = (1 − 0.27)4 = 0. 734 = 0.2840 Method 2: Direct, but Tedious Method Using Compliment Rule ( ≥ 5) = 1 − ( < 5) = 1 − (( = 1) + ( = 2) + ( = 3) + ( = 4)) = 1 − ( = 1) − ( = 2) − ( = 3) − ( = 4) = 1 − (1 − ) − (1 − )2−1 − (1 − )3−1 − (1 − )4−1 = 1 − 0.27 − 0.27 ∗ 0. 731 − 0.27 ∗ 0. 732 − 0.27 ∗ 0.733 = 0.2840 d) Find the probability he surveys between 6 people and 8 people (inclusive) until he finds one who attended a PMO performance last semester. (6 ≤ ≤ 8) = ( = 6) + ( = 7) + ( = 8) = (1 − )6−1 + (1 − )7−1 + (1 − )8−1 = 0.27 ∗ (0.73)5 + 0.27 ∗ (0.73)6 + 0.27 ∗ (0.73)7 = 0.1267 1−1 e) He has already surveyed 7 people and has not found any who attended a PMO performance last semester. What is the probability it takes him at most ten people to find one who attended a PMO performance last semester? There are two methods of calculating the desired probability. The first uses the so-called tail property for Geometric distribution. The second method is more direct, but laborious way. Method 1: Memoryless Property along with complement rule and tail property ( ≤ 10| > 7) = ( ≤ 3) = 1 − ( > 3) = 1 − (1 − )3 = 1 − (1 − 0.27)3 = 0.6110 Method 2: Direct, but laborious way ( ≤ 10 ∩ > 7) (7 < ≤ 10) ( = 8) + ( = 9) + ( = 10) ( ≤ 10| > 7) = = = ( > 7) ( > 7) ( > 7) 8−1 9−1 10−1 (1 − ) + (1 − ) + (1 − ) = (1 − )7 0.27 ∗ (0.73)7 + 0.27 ∗ (0.73)8 + 0.27 ∗ (0.73)9 = = 0.6110 0. 737 f) The representative needs to find 10 people who have attended a PMO performance last semester. How many should he expect to survey in order to find 10 people who attended a performance last semester? Also find the standard deviation of this value. As we now must find 10 people, we have a new random variable. Denote this as N. ~( = 10, = 0.27) 10 [] = = .27 = 37.0370 () = √(1 − )/2 = √10(1 − 0.27)/0. 272 = √10 ∗ 0.73/0. 272 = 10.0069 4. Textbook, Exercise 19.6 (Cat fur vaccine), page 250 a) Explain in words what X is in terms of the story. What values can it take? ℎ ℎ 50 ( 4000 )ℎ ℎ , ℎ ℎ 1250 ℎ . 0, 1, 50. b) Why is this a Hypergeometric situation? What are the parameters? ℎ ℎ ℎ ℎ , . . ℎ ℎ ( ℎ 4000 , ℎ 1250 ℎ ). ℎ = 4000, = 50, = 1250. c) What are the expected number of people in this sample who have cat fur allergies? 1250 [] = ∗ = 50 ∗ = 15.625 4000 d) What is the probability that exactly 20 of the people in his sample have cat fur allergies? Just set up the equation, but don’t try to solve it? ( = 20) = ( 1250 2750 )( ) 20 30 4000 ( ) 50 e) What would be a good approximation to use in this situation? What are the parameters? = 4000, = 50, > 20 ∗ 4000 > 1000 ℎ . 1250 ℎ = 50 = = 0.3125 4000 f) What is the approximate probability that 20 of the people in his sample have cat fur allergies? Set up the equation and solve. 50 ( ∗ = 20) = ( ) 0. 312520 ∗ 0.687530 = 0.0488 20 5. Textbook, Exercise 21.10-SKIP PART C!!! (Chinese checkers), page 263-264 a) In a tournament, Philip and Callum will play 8 games. What is the probability Philip wins exactly 6 games? X is number of games that Philip wins of the 8 played 8 ~( = 8, = 0.7) ( = 6) = ( ) 0. 76 ∗ 0. 32 = 0.2965 6 b) Philip and Callum decide to keep playing until Callum wins 3 times. What is the probability they play a total of 12 games? Y is the number of games played until Callum wins his 3rd game. ~( = 3, = 0.3) 11 ( = 12) = ( ) 0. 33 ∗ 0. 79 = 0.0599 2 6. Textbook, Exercise 22.12 (Babies), page 264 a) What is the probability that at least one baby will be born today on the 8 am to 4 pm shift? Denote the number babies born from 8 am – 4 pm as A. 9 ~ ( = ∗ 8 ℎ = 3) 24 ℎ − 0 ( ≥ 1) = 1 − ( = 0) = 1 − = 1 − −3 = 0.9502 0! b) What is the probability that at least one baby will be born in the next hour? 9 ~( = = 0.375) 24 − 0 ( ≥ 1) = 1 − ( = 0) = 1 − = 1 − −9/24 = 0.3127 0! c) What is the probability that exactly 4 babies will be born on the 8 am to 4 pm shift? − 4 −3 34 ( = 4) = = = 0.1680 4! 4! d) What is the probability that exactly 4 babies will be born on each of the next three 8 hours shifts? Let 1 , 2 , 3 denote the next three 8-hour shifts each of these is ( = 9 24 ℎ ∗ 8 ℎ = 3) 3 −3 34 −3 34 −3 34 −3 34 (1 = 4 ∩ 2 = 4 ∩ 3 = 4) = (1 = 4)(2 = 4)(3 = 4) = ∗ ∗ =[ 4! 4! 4! 4! = 0.0047 e) What is the probability that you would have to wait for four 8 hour shifts until you got the first one with exactly 4 babies born? This question is nested. Define W as the number of 8-hr shifts until the first one with exactly 4 babies born. ~( = ( = 4) = 0.1680) see answer to part c) ( = 4) = (1 − )4−1 = 0.1680 ∗ (1 − 0.1680)3 = 0.0968 f) What is the probability that exactly 12 babies will be born total in the next 24 hours? ~ ( = ( = 12) = 9 ∗ 24 ℎ = 9) 24 ℎ − 12 −9 912 = = 0.0728 12! 12! 7. Textbook, Exercise 21.18 (Guessing on an exam), page 266 a) What is the probability that a person who randomly guesses on each question gets exactly 5 questions correct given that they got at least 1 question correct? Let A denote the number of questions correct. ( = 5| ≥ 1) = 1 ~( = 20, = 5 = 0.2) (20 )0. 25 ∗ 0. 815 ( = 5 ∩ ≥ 1) ( = 5) ( = 5) 5 = = = = 0.1766 ( ≥ 1) ( ≥ 1) 1 − ( = 0) 1 − (20 )0. 20 ∗ 0. 820 0 b) What is the expected number of correct answers on the exam? What is the standard deviation? [] = = 20 ∗ 0.2 = 4 () = √(1 − ) = √20 ∗ 0.2 ∗ 0.8 = 1.7889 c) What is the probability I will have to grade more than 12 exams to find one with exactly 5 correct answers, assuming all of my students were randomly guessing on all the questions? ~( = ( = 5) = (20 )0. 25 ∗ 0. 815 = 0.1746) 5 This question is nested We use the tail property. ( > 12) = (1 − )12 = (1 − 0.1746)12 = 0.099994 ≈ 0.1000 d) What is the expected number of correct answers if a person takes 6 exams? What is the standard deviation in the number of correct answers if a person takes 6 exams? There are two interpretations to this question. We will accept either. The first is that we simply multiply our random variable by 6. The second is we have 6 different exams created a binomial with = 120. First interpretation: [6] = 6[] = 6 = 6 ∗ 20 ∗ 0.2 = 24 (6) = |6|() = 6() = 6√(1 − ) = 6√20 ∗ 0.2 ∗ 0.8 = 10.7331 Second Interpretation ~( = 120, = .2) [] = = 120 ∗ 0.2 = 24 () = √120 ∗ 0.2 ∗ 0.8 = 4.3818 e) If you have to pay $2 to take the exam, but your parents pay you 50 cents for every correct answer, what is your expected net profit? What is the standard deviation in your net profit? = − = 0.5 ∗ − 2 [] = [0.5 − 2] = 0.5[] − 2 = 0.5 ∗ 20 ∗ 0.2 − 2 = 0 () = √(0.5 − 2) = |. 5|√20 ∗ 0.2 ∗ 0.8 = 0.8944 8. Textbook, Exercise 21.20 (Coffee beans), page 267 a) What is the expected number of incorrectly roasted beans in the 1000-item sample? What is the standard deviation? Define as the number of incorrectly roasted beans identified by the employees. ~( = 1000, = 0.008) [] = = 1000 ∗ (0.008) = 8 [] = √(1 − ) = √1000 ∗ 0.008 ∗ (1 − 0.008) = 2.8171 b) Set up the exact probability (but do not solve) that at least 2 beans from the sample are incorrectly roasted. ( ≥ 2) = 1 − ( = 0) − ( = 1) = 1 − (1000 )0. 0080 (1 − 0.008)1000 − (1000 )0.008(1 − 0.008)999 0 1 c) What is the approximate probability that at least 2 beans from the sample are incorrectly roasted? We can use the Poisson Approximation to the Binomial. Observe = 0.008 < 0.01 and = 1000 > 100 Then ∗ ~( = = 1000 ∗ 0.008 = 8) (∗ ≥ 2) = 1 − (∗ = 0) − (∗ = 1) = 1 − − 0 − 1 − = 1 − −8 − 8 −8 = 0.9970 0! 1! ...
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