Recitation 22 Problem 1. Final exams in 6.042 are graded according to a rigorous procedure With probability the exam is graded by a recitation instructor with probability, it is graded by a lecturer, and with probability ,, it is accidentally dropped behind the radiator and arbitrarily given a score of 84 Recitation instructors score an exam by scoring each problem individually and then taking th ne sum There are ten true/false questions worth 2 points each. For each, full credit given with probability ,, and no credit is given with probability There are four questions worth 15 points each. For each, the score is deter- mined by rolling two fair dice, summing the results, and adding 3 The single 20 point question is awarded either 12 or 18 points with equal prob- Lecturers score an exam by rolling a fair die twice, multiplying the results, and then adding a general impression"score With probability io, the general impression score is 40 With probability ao, the general impression score With probability io, the general impression score Assume all random choices during the grading process are mutually independent (a)What is the expected score on an exam graded by a recitation instructor? Solution. Let X equal the exam score and C be the event that the exam is graded by a recitation instructor. We want to calculate Ex(X C) By linearity of (conditional) expectation, the expected sum of the problem scores is the sum of the expected problem scores. Therefore, we have Ex(X I C)=10. Ex(T/F score C)+4. Ex(15pt prob score |C)+Ex(20pt prob score I C 7 1 +-0+4·2·+3)+ 12+-·18 =102+4:10+15=70 (b) What is the expected score on an exam graded by a lecturer?Recitation 22 4 Problem 1. Final exams in 6.042 are graded according to a rigorous procedure: • With probability 4 7 the exam is graded by a recitation instructor, with probability 2 it 7 is graded by a lecturer, and with probability 1 7 , it is accidentally dropped behind the radiator and arbitrarily given a score of 84. • Recitation instructors score an exam by scoring each problem individually and then taking the sum. – There are ten true/false questions worth 2 points each. For each, full credit is given with probability 3 4 , and no credit is given with probability 1 . 4 – There are four questions worth 15 points each. For each, the score is deter￾mined by rolling two fair dice, summing the results, and adding 3. – The single 20 point question is awarded either 12 or 18 points with equal prob￾ability. • Lecturers score an exam by rolling a fair die twice, multiplying the results, and then adding a “general impression” score. 4 – With probability 10 , the general impression score is 40. 3 – With probability 10 , the general impression score is 50. 3 – With probability 10 , the general impression score is 60. Assume all random choices during the grading process are mutually independent. (a) What is the expected score on an exam graded by a recitation instructor? Solution. Let X equal the exam score and C be the event that the exam is graded by a recitation instructor. We want to calculate Ex (X | C). By linearity of (conditional) expectation, the expected sum of the problem scores is the sum of the expected problem scores. Therefore, we have: Ex (X C| ) = 10 · Ex (T/F score C) + 4 · Ex (15pt prob score | C) + Ex (20pt prob score C) � � | � � � � | 3 1 7 1 1 = 10 · 2 + 0 + 4 · 2 · + 3 + 12 + 18 4 · 4 · 2 2 · 2 · 3 = 10 · + 4 · 10 + 15 = 70 2 (b) What is the expected score on an exam graded by a lecturer?