{"id":97617,"date":"2022-05-15T15:24:11","date_gmt":"2022-05-15T09:54:11","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=97617"},"modified":"2022-05-10T14:22:29","modified_gmt":"2022-05-10T08:52:29","slug":"eureka-math-precalculus-module-5-lesson-8","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-precalculus-module-5-lesson-8\/","title":{"rendered":"Eureka Math Precalculus Module 5 Lesson 8 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math Precalculus Module 5 Lesson 8 Answer Key<\/h2>\n

Eureka Math Precalculus Module 5 Lesson 8 Exploratory Challenge \/Exercise Answer Key<\/h3>\n

Exploratory Challenge 1\/Exercise 1
\nRecall the following problem from the Problem Set in Lesson 7:
\nSuppose two dice are rolled. The sum of the two numbers showing is a discrete random variable. The following table displays the probability distribution of this random variable:
\n\"Engage
\nExercise 1.
\nIf you rolled two dice and added the numbers showing a large number of times, what would you expect the average sum to be? Explain why.
\nAnswer:
\nThe expected sum of the two rolled dice is as follows:
\n2(\\(\\frac{1}{36}\\)) + 3(\\(\\frac{1}{18}\\)) + 4(\\(\\frac{1}{12}\\)) + 5(\\(\\frac{1}{9}\\)) + 6(\\(\\frac{5}{36}\\)) + 7(\\(\\frac{1}{6}\\)) + 8(\\(\\frac{5}{36}\\)) + 9(\\(\\frac{1}{9}\\)) + 10(\\(\\frac{1}{12}\\)) + 11(\\(\\frac{1}{18}\\)) + 12(\\(\\frac{1}{36}\\)) = 7
\nThe expected average sum would be 7 because, after a large number of rolls, the distribution of sums would resemble the probability distribution above.<\/p>\n

Exploratory Challenge 1\/Exercises 2\u20134
\nExercise 2.
\nRoll two dice. Record the sum of the numbers on the two dice in the table below. Repeat this nine more times for a total of 10 rolls.
\n\"Engage
\nAnswer:
\nStudent responses will vary. Here is one example.
\n\"Engage<\/p>\n

Exercise 3.
\nWhat is the average sum of these 10 rolls?
\nAnswer:
\nAnswers will vary. For the example above, the average of the sum for these 10 rolls is as follows:
\n3(0.1) + 4(0.1) + 5(0.2) + 6(0.2) + 7(0.1) + 9(0.2) + 10(0.1) = 6.4<\/p>\n

Exercise 4.
\nHow does this average compare to the expected value in Exercise 1? Are you surprised? Why or why not?
\nAnswer:
\nAnswers will vary, but for most students, the average will not be the same as the expected sum of 7 in Exercise 1. Students may say they are not surprised as there were only 10 rolls of the dice.<\/p>\n

Exploratory Challenge 1\/Exercises 5\u20137
\nExercise 5.
\nRoll the two dice 10 more times, recording the sums. Combine the sums of these 10 rolls with the sums of the previous 10 rolls for a total of 20 sums.
\n\"Engage
\nAnswer:
\nStudent responses will vary. Here is one example.
\n\"Engage<\/p>\n

Exercise 6.
\nWhat is the average sum for these 20 rolls?
\nAnswer:
\nAnswers will vary. For the example above, the average sum for these 20 rolls is as follows:
\n3(0.15) + 4(0.05) + 5(0.15) + 6(0.25) + 7(0.20) + 9(0.10) + 10(0.10) = 6.2<\/p>\n

Exercise 7.
\nHow does the average sum for these 20 rolls compare to the expected value in Exercise 1?
\nAnswer:
\nAnswers will vary, but for most students, the average will still not be the same as the expected sum of 7 in Exercise 1.<\/p>\n

Exploratory Challenge 1\/Exercise 8
\nExercise 8.
\nCombine the sums of your 20 rolls with those of your partner. Find the average of the sum for these 40 rolls.
\n\"Engage
\nAnswer:
\nStudent responses will vary. Here is one example.
\n\"Engage
\nThe average sum for these 40 rolls is as follows:
\n3(0.075) + 4(0.075) + 5(0.15) + 6(0.2) + 7(0.125) + 8(1.25) + 9(0.10) + 10(0.10) + 11(0.025) + 12(0.025)
\n= 6.825<\/p>\n

Exploratory Challenge 2\/Exercise 9
\nExercise 9.
\nCombine the sums of your 40 rolls above with those of another pair for a total of 80 rolls. Find the average value of the sum for these 80 rolls.
\nAnswer:
\nStudent responses will vary. Here is one example.
\n\"Engage
\nThe average sum for these 80 rolls is as follows:
\n3(0.075) + 4(0.05) + 5(0.1625) + 6(0.175) + 7(0.1375) + 8(0.1125) + 9(0.075) + 10(0.15) + 11(0.0375) + 12(0.025) = 7.0375<\/p>\n

Exploratory Challenge 2\/Exercises 10\u201311
\nExercise 10.
\nCombine the sums of your 80 rolls with those of the rest of the class. Find the average sum for all the rolls.
\n\"Engage
\nAnswer:
\nStudent responses will vary. Here is one example for a class of 20 students.
\n\"Engage
\nThe average sum of these 400 rolls is as follows:
\n2(0.0175) + 3(0.0675) + 4(0.0825) + 5(0.12) + 6(0.15) + 7(0.155) + 8(0.1175) + 9(0.1075) + 10(0.0975) + 11(0.0575) + 12(0.0275) = 6.9975<\/p>\n

Exercise 11.
\nThink about your answer to Exercise 1. What do you notice about the averages you have calculated as the number of rolls increases? Explain why this happens.
\nAnswer:
\nAs the number of rolls increases, the average value approaches the expected value. This happens because as the number of observed values of the random variable increases, the relative frequencies become closer to the actual probabilities in the probability distribution. The expected value of a discrete random variable is a long-run average value for the variable.<\/p>\n

Exploratory Challenge 2\/Exercise 12
\nThe expected value of a discrete random variable is the long-run mean value of the discrete random variable. Refer back to Exercise 1 where two dice were rolled, and the sum of the two dice was recorded. The interpretation of the expected value of a sum of 7 would be
\nWhen two dice are rolled over and over for a long time, the mean sum of the two dice is 7.
\nNotice that the interpretation includes the context of the problem, which is the random variable sum of two dice, and also includes the concept of long-run average.<\/p>\n

Exercise 12.
\nSuppose a cancer charity in a large city wanted to obtain donations to send children with cancer to a circus appearing in the city. Volunteers were asked to call residents from the city\u2019s telephone book and to request a donation. Volunteers would try each phone number twice (at different times of the day). If there was no answer, then a donation of $0 was recorded. Residents who declined to donate were also recorded as $0. The table below displays the results of the donation drive.
\n\"Engage
\nFind the expected value for the amount donated, and write an interpretation of the expected value in context.
\nAnswer:
\nThe expected amount donated is as follows:
\n0(0.11) + 10(0.35) + 20(0.25) + 50(0.20) + 100(0.09) = 27.50
\nWhen a large number of residents are contacted, the mean amount donated is $27.50.<\/p>\n

Eureka Math Precalculus Module 5 Lesson 8 Problem Set Answer Key<\/h3>\n

Question 1.
\nSuppose that a discrete random variable is the number of broken eggs in a randomly selected carton of one dozen eggs. The expected value for the number of broken eggs is 0.48 eggs. Which of the following statements is a correct interpretation of this expected value? Explain why the others are wrong.
\na. The probability that an egg will break in one-dozen cartons is 0.48, on average.
\nb. When a large number of one-dozen cartons of eggs are examined, the average number of broken eggs in a one-dozen carton is 0.48 eggs.
\nc. The mean number of broken eggs in one-dozen cartons is 0.48 eggs.
\nAnswer:
\nThe correct answer is b.
\nAnswer choice a relates the expected value to a probability, which is incorrect. The expected value is the long-run mean of a random variable.
\nAnswer choice c does not refer to the long-run aspect of the expected value.<\/p>\n

Question 2.
\nDue to state funding, attendance is mandatory for students registered at a large community college. Students cannot miss more than eight days of class before being withdrawn from a course. The number of days a student is absent is a discrete random variable. The expected value of this random variable for students at this college is 3.5 days. Write an interpretation of this expected value.
\nAnswer:
\nIf many students at this college are asked how many days they have been absent, the average number would be 3.5 days.<\/p>\n

Question 3.
\nThe students at a large high school were asked to respond anonymously to the question:
\nHow many speeding tickets have you received?
\nThe table below displays the distribution of the number of speeding tickets received by students at this high school.
\n\"Eureka
\nCompute the expected number of speeding tickets received. Interpret this mean in context.
\nAnswer:
\nThe expected number of speeding tickets received by students at this high school is as follows:
\n0(0.55) + 1(0.28) + 2(0.09) + 3(0.04) + 4(0.03) + 5(0.01) = 0.75
\nIn the long run, students at this high school have received an average of 0.75 speeding tickets.<\/p>\n

Question 4.
\nEmployees at a large company were asked to respond to the question:
\nHow many times do you bring your lunch to work each week?
\nThe table below displays the distribution of the number of times lunch was brought to work each week by employees at this company.
\n\"Eureka
\nCompute the expected number of times lunch was brought to work each week. Interpret this mean in context.
\nAnswer:
\n0(0.30) + 1(0.12) + 2(0.12) + 3(0.10) + 4(0.06) + 5(0.30) = 2.4
\nIn the long run, employees at this company brought lunch, on average, 2.4 times each week.<\/p>\n

Question 5.
\nGraduates from a large high school were asked the following:
\nHow many total AP courses did you take from Grade 9 through Grade 12?
\nThe table below displays the distribution of the total number of AP courses taken by graduates while attending this high school.
\n\"Eureka
\nCompute the expected number of total AP courses taken per graduate. Interpret this mean in context.
\nAnswer:
\nThe expected number of AP courses taken is as follows:
\n0(0.575) + 1(0.06) + 2(0.09) + 3(0.12) + 4(0.04) + 5(0.05) + 6(0.035) + 7(0.025) + 8(0.005) = 1.435
\nIf many high school graduates are asked how many AP courses were taken between Grades 9 and 12, the average number would be 1.435 AP courses.<\/p>\n

Question 6.
\nAt an inspection center in a large city, the tires on the vehicles are checked for damage. The number of damaged tires is a discrete random variable. Create two different distributions for this random variable that have the same expected number of damaged tires. What is the expected number of damaged tires for the two distributions? Interpret the expected value.
\nAnswer:
\nAnswers will vary. Here is one example.<\/p>\n

Distribution 1:
\n\"Eureka
\nAnswer:
\n\"Eureka
\nThe expected number of damaged tires is as follows:
\n0(0.60) + 1(0.20) + 2(0.10) + 3(0.05) + 4(0.05) = 0.75
\nThe expected number of damaged tires is 0.75.<\/p>\n

Distribution 2:
\n\"Eureka
\nAnswer:
\n\"Eureka
\nThe expected number of damaged tires is as follows:
\n0(0.55) + 1(0.30) + 2(0.05) + 3(0.05) + 4(0.05) = 0.75
\nBecause this inspection center examines the tires on the vehicles of a large number of customers, the inspectors find an average of 0.75 damaged tires per vehicle.<\/p>\n

Eureka Math Precalculus Module 5 Lesson 8 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nAt a large university, students are allowed to register for no more than 7 classes. The number of classes for which a student is registered is a discrete random variable. The expected value of this random variable for students at this university is 4.15.
\nWrite an interpretation of this expected value.
\nAnswer:
\nIf many students at this university are asked how many classes they are registered for, the average number would be 4.15 classes.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math Precalculus Module 5 Lesson 8 Answer Key Eureka Math Precalculus Module 5 Lesson 8 Exploratory Challenge \/Exercise Answer Key Exploratory Challenge 1\/Exercise 1 Recall the following problem from the Problem Set in Lesson 7: Suppose two dice are rolled. The sum of the two numbers showing is a discrete random variable. … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/97617"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=97617"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/97617\/revisions"}],"predecessor-version":[{"id":136343,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/97617\/revisions\/136343"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=97617"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=97617"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=97617"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}