2024 Test 1 Form B Solutions

AMS 310 Spring 2024 Test 1 Form B Solutions Fred Rispoli 1. The following side-by-side boxplots display the first and second midterm scores of an introductory statistics course. a) Which midterm has a larger standard deviation of scores? b) Overall, did students perform better or worse on the second midterm compared to the first midterm. c) On the second midterm what can be said about skewness (i.e. is there very little skewness or significant skewness?) d) Use the boxplot for midterm 1 and give the 5 number summary. a) (2) Midterm 1, the box plot is more spread out b) (2) Better on Midterm 2, students had a higher mean score and a higher median score c) (3) There is little skewness, the x and the horizontal line are very close d) (3) min = 20 or 30, Q1 = 59 or 60, median = 75, Q3 = 85, max = 100 2. Consider the following data set which gives weights of 12 young goats, in pounds: 52, 32, 60, 59, 74, 65, 44, 51, 58, 51, 68, 47 a) Find the sample variance. Show steps used. b) Find the 40th percentile. n a) (5) Use 2 i n ( ) ∑ x − ∑ xi 2 s= i=1 2 i=1 n−1 /n = 37845−43,692.1 = 130.447 11 b) (5) First sort: 32, 44, 47, 51, 51, 52, 58, 59, 60, 65, 68, 74 This study source was downloaded by 100000819498514 from CourseHero.com on 11-07-2024 09:43:46 GMT -06:00 https://www.coursehero.com/file/227432015/AMS-310-2024-Test-1-Form-B-Solutions-docx/ np = 12 (0.40) = 4.8, use the 5th value which is 51 3. Below is a Pareto Chart for complaints at a certain restaurant. The total number of complaints is 1,722. a) Which categories of complaints account for roughly 80% of all complaints? b) Suppose a new complaint is recorded. Approximate the probability that the new complaint is due to small portions or wait time. Round to two decimal places. a) (5) Overpriced and Small portions b) (5) Prob = 109+ 621 = 0.42 1722 4. Hot dog eating contests are popular at the famous Nathans in Brooklyn. The table below gives information about BMI and age group for a group of adults in a hot dog eating contest. Age 18 to 30 31 to 50 51 to 65 66 to 80 Total Low BMI 0.09 0.11 0.10 0.04 0.34 Normal BMI 0.06 0.18 0.12 0.07 0.43 High BMI 0.05 0.10 0.06 0.02 0.23 Total 0.20 0.39 0.28 0.13 1.00 a) A person selected at random from this group is found to be normal BMI. What is the probability that this person is between 31 to 50 years old? Round to two decimal places. b) A person selected at random from this group. What is the probability that the person has a normal BMI, given that their is age 51 to 65? Round to two decimal places. This study source was downloaded by 100000819498514 from CourseHero.com on 11-07-2024 09:43:46 GMT -06:00 https://www.coursehero.com/file/227432015/AMS-310-2024-Test-1-Form-B-Solutions-docx/ c) A person is selected at random from this group. Given that the person's BMI is not high, what is the probability their age is between 66 and 80? a) (3) P(between 31 and 50 │Normal BMI) = b) (3) P(Normal BMI │ between 51 to 65) = 0.18 = 0.42 0.43 0.12 = 0.43 0.28 c) (4) P(between 66 to 80│ BMI is not high) = 0. 04+0.07 = 0.14 0. 34+0.43 5. A and B are events such that P(A) = 0.4 and P(A ∪ B) = 0.6. a) Find P(B) assuming that A and B are mutually exclusive. b) Find P(B) assuming that A and B are independent. c) P(A │ B) = 0.2 (a) (3) 0.6=P ( A ∪ B )=P ( A ) + P ( B )=0.4+ P ( B ) ⇒ P ( B )=0.6−0.4=0.2 (b) (4) 0.6=P ( A ∪ B )=P ( A ) + P ( B )−P ( A ∩ B )=P ( A ) + P ( B ) −P ( A ) P ( B ) ¿ 0.4+ P ( B )−0.4 P ( B ) ⇒ P ( B )−0.4 P ( B )=0.6−0.4 ⇒ 0.6 P ( B )=0.2 ⇒ 0.2 1 P ( B )= = = 0.33 0.6 3 P ( A ∩ B ) P ( A )+ P ( B )−P ( A ∪ B ) 0.4+ P ( B )−0.6 = = P (B) P( B) P(B) 0.2 1 ⇒ 0.2 P ( B )=P ( B )−0.2 ⇒ 0.8 P ( B )=0.2 ⇒ P ( B )= = = 0.25 0.8 4 6. Suppose two different methods are available for eye surgery. The probability that the eye has not recovered in a month is 0.005 if method A is used. When method B is used, the probability that the eye has not recovered in a month is 0.007. Assume that 45% of eye surgeries are done with method A and 55% are done with method B in a certain hospital. If an eye is recovered in a month after surgery is done in the hospital, what is the probability that method B was performed (give an answer with three decimal places)? Show your reasoning. c) (3) 0.2=P ( A∨B )= Use Bayes Theorem: Let A = method A is used; Let B = method B is used Let R = an eye is recovered in a month after surgery. Find P(B|R) P(A) = 0.45, P(B) = 0.55, P(R|A) = 0.995, P(R|B) = 0.993, P(R) = P(R|A)P(A) + P(R|B)P(B), so P(R) = (.995)(.45) + (.993)(.55) = 0.44775 + .54615 = 0.9939 P ( B|R )= P(B∩ R) P ( B ) ( 0.995 ) ( 0. 55 ) = ≈ 0.5 50 0.9939 P(R) This study source was downloaded by 100000819498514 from CourseHero.com on 11-07-2024 09:43:46 GMT -06:00 https://www.coursehero.com/file/227432015/AMS-310-2024-Test-1-Form-B-Solutions-docx/ 7. Famous WNBA star Sue Bird hits 90% of her free throws in basketball games in 2022. Suppose that in a certain 2022 basketball game she had 20 free throws. a) What is the probability that she made at least 18 free throws? Round your answer to 3 decimal places. b) What is the probability that she makes less than 20 free throws? Round your answer to 3 decimal places. c) Suppose that in the year 2020 she hit 80% of her free throws. In a game with 20 free throws how many should we expect her to make and what is the standard deviation? Students must use formulas here. a) (4) P(X ≥ 18) = P(18) + P(19) + P(20) = 0.285 + 0.27017 + 0.12158 = 0.667 b) (3) P(X < 20) = 1 – P(20) = 1 - 0.122 = 0.878 c) (3) µ = np = 20(.80) = 16  = sqrt[(20)(.8)(.2)] = 1.79 8. Suppose a representative at a credit card customer service center receives 5 phone calls every minute on average. a. Find the probability that she receives 10 phone calls in 3 minutes. Round off your answer to three decimal places. b. Find the probability that she receives at least 5 phone calls in 2 minutes. Round off your answer to three decimal places. c. Find the probability that she receives at most 2 phone calls in 1 minute. Round off your answer to three decimal places. Students must use formulas here. a) (3) First note that we must use the Poisson distribution with  = 15 10 −15 P(X =10) = 15 e 10 ! = 0.049 b) (4) First note that  = 10 P(X ≥ 5) = 1 - P(0) - P(1) - P(2) – P(3) – P(4) = 0.971 c) (3) Now  = 1 P(X ≤ 2) = P(0) + P(1) + P(2) = 0.125 This study source was downloaded by 100000819498514 from CourseHero.com on 11-07-2024 09:43:46 GMT -06:00 https://www.coursehero.com/file/227432015/AMS-310-2024-Test-1-Form-B-Solutions-docx/ 9. a) A company in New Jersey is hiring 10 people for a position. Of the applicants, 25 are from New Jersey, 15 are from New York. If the applicants are chosen at random from the pool of 40 applicant, what is the probability that of the 10 people hired exactly 5 are from New York? Explain what distribution is needed to find the probability and give an answer rounded off to 3 decimal places. b) Lisa is rolling two dice. She keeps rolling until she gets doubles (i.e., the same number on both dice.) What is the probability she obtains her first doubles on the third roll or fourth roll? Explain what distribution is needed to find the probability and give an answer rounded off to 3 decimal places. 25 15 ( 5 )( 5 ) =¿ ¿ a) Use the hypergeometric distribution. Prob = 40 ( 10 ) b) Use Geometric Distribution The Probability that she rolls doubles: p = 1 6 P(X = 3) + P(X = 4) = (5/6)2(1/6) + (5/6)3(1/6) = 0.116 + 0.097 = 0.212 10. a) For an arbitrary data set with mean of 151 and standard deviation of 14, at least what percent of the values will fall between 95 and 207? Round your answer to 4 decimal places. b) Suppose that one wanted to be sure that at least 85% of an arbitrary data set with mean 100 and standard deviation 15 lies within a certain interval. What must the interval be? Round off to one decimal place. a) The values between 95 and 207 are all within 56 units of the mean. Since 56 = 4(14). Let k = 4 and use Chebyshev's formula to find the fraction of the data that are within k=4 1 standard deviations of the mean. We get 1 - 2 = 15/16 or 0.9375%. k b) Set 1 - 1 = 0.85 and solve for k. This gives k = √ 6.66 ≈ 2.58. k2 100 + 2.58(15) = 100 + 38.7 = 138.7 and 100 – 2.58(15) = 61.3 The interval is 61.3 to 138.7. This study source was downloaded by 100000819498514 from CourseHero.com on 11-07-2024 09:43:46 GMT -06:00 https://www.coursehero.com/file/227432015/AMS-310-2024-Test-1-Form-B-Solutions-docx/ Powered by TCPDF (www.tcpdf.org)