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Approximately 3.5 percent of all children born in a certain region are from multiple births (that is, twins, triplets, etc.). Of the children born in the region who are from multiple births, 22 percent are left-handed. Of the children born in the region who are from single births, 11 percent are left-handed. (a) What is the probability that a randomly selected child born in the region is left-handed? (b) What is the probability that a randomly selected child born in the region is a child from a multiple birth. given that the child selected is left-handed? (c) A random sample of 20 children born in the region will be selected. What is the probability that the sample will have at least 3 children who are left-handed? 1. What was your score for part (a) (E, P or I). Explain how you met the scoring components or what you should have done differently. 2. What was your score for part (b) (E, P or I). Explain how you met the scoring components or what you should have done differently. 3. What was your score for part (c) (E, P or I). Explain how you met the scoring components or what you should have done differently. 4. What was your overall score on this problem? Write a statement about what you did correctly, what you did incorrectly and anything you should take note of when completing problems like these.
Approximately 3.5 percent of all children born in a certain region are from multiple births (that is, twins, triplets, etc.). Of the children born in the region who are from multiple births, 22 percent are left-handed. Of the children born in the region who are from single births, 11 percent are left-handed. (a) What is the probability that a randomly selected child born in the region is left-handed? (b) What is the probability that a randomly selected child born in the region is a child from a multiple birth. given that the child selected is left-handed? (c) A random sample of 20 children born in the region will be selected. What is the probability that the sample will have at least 3 children who are left-handed? 1. What was your score for part (a) (E, P or I). Explain how you met the scoring components or what you should have done differently. 2. What was your score for part (b) (E, P or I). Explain how you met the scoring components or what you should have done differently. 3. What was your score for part (c) (E, P or I). Explain how you met the scoring components or what you should have done differently. 4. What was your overall score on this problem? Write a statement about what you did correctly, what you did incorrectly and anything you should take note of when completing problems like these.
Final answer:The probability that a randomly selected child born in the region is left-handed is 11.39%. The probability that a randomly selected child born in the region is a child from a multiple birth, given that the child selected is left-handed, is 1.93%. The probability that a sample of 20 children born in the region will have at least 3 children who are left-handed is 54.37%.Explanation:(a) To find the probability that a randomly selected child born in the region is left-handed, we need to consider both multiple births and single births. The probability can be calculated as follows:Probability of selecting a child from multiple births and left-handed = (3.5%)(22%) = 0.77%Probability of selecting a child from single births and left-handed = (96.5%)(11%) = 10.62%Summing up these probabilities: 0.77% + 10.62% = 11.39%Therefore, the probability that a randomly selected child born in the region is left-handed is 11.39%.(b) To find the probability that a randomly selected child born in the region is a child from a multiple birth, given that the child selected is left-handed, we can use conditional probability. The probability can be calculated as follows:Probability of selecting a left-handed child from multiple births = 22% (given in the question)Probability of selecting a left-handed child = 11.39% (calculated in part a)Therefore, the probability that a randomly selected child born in the region is a child from a multiple birth, given that the child selected is left-handed, is (22%)/(11.39%) = 1.93%.(c) To find the probability that a sample of 20 children born in the region will have at least 3 children who are left-handed, we can use binomial probability. The probability can be calculated as follows:Probability of selecting at least 3 left-handed children in a sample of 20 = 1 - (probability of selecting 0 or 1 left-handed child in a sample of 20)Probability of selecting 0 left-handed child in a sample of 20 = (88.61%)^20 = 11.72%Probability of selecting 1 left-handed child in a sample of 20 = (20C1)(88.61%)^19(11.39%) = 33.91%Summing up these probabilities: 1 - (11.72% + 33.91%) = 54.37%Therefore, the probability that a sample of 20 children born in the region will have at least 3 children who are left-handed is 54.37%.Learn more about Probability here:brainly.com/question/22962752#SPJ11...