2023-10-06 Test1S B11

Page 3 of 9 Important note: Put your answers of the multiple choice questions in Page 2 of this test paper. Only Page 2 is marked for multiple choice questions. Part 1. Multiple Choice. Select one answer per question. Total: 17 questions = 51 points. 1. A random sample of 5 students has the following test scores 50, 60, 60, 65, 70. The 50 th percentile is closest to (a) 50 √ (b) 60 (c) 65 (d) 68 (e) 70 2. A random sample of 5 students has the following test scores 50, 60, 60, 65, 70. The standard deviation is closest to √ (a) 7.42 (b) 9.45 (c) 12.33 (d) 15.36 (e) 20.12 3. A group of 50 students consists of 30 girls and 20 boys. The median test score for the girls is 80 and the median test score for the boys is 70. Which one of the following statements is most correct on the median of these 50 students? (a) Nothing can be known about the value of the median since the 50 test scores are not given. (b) Median = (80 + 70)/2 = 75 (c) Median = (30 × 80 + 20 × 70)/50 = 76 (d) Median = 150 √ (e) 70 ≤ Median ≤ 80 4. A sample of 100 students has the following test scores: 50, 60, …, 70. Given that the following calculations: 50 + 60 + ⋯ + 70 = 600 and 50 2 + 60 2 + ⋯ + 70 2 = 25875 . The sample standard deviation is closest to (a) 5 (b) 10 (c) 12.5 √ (d) 15 (e) 20 5. A simple regression model is assumed for relating the price of grapefruit X (in dollars) to quantities of grapefruit demanded Y . Data for five months is provided below and some calculations are given. X Y ? − ? ̅ (? − ? ̅ ) 2 ? − ? ̅ (? − ? ̅ ) 2 (? − ? ̅ )(? − ? ̅ ) 0.1 50 – 0.2 0.04 22 484 – 4.4 0.2 30 – 0.1 0.01 2 4 – 0.2 0.3 30 0 0 2 4 0 0.4 20 0.1 0.01 – 8 64 – 0.8 0.5 10 0.2 0.04 – 18 324 – 3.6 Total 1.5 140 0 0.10 0 880 – 9.0 Mean 0.3 28 0 0.02 0 176 – 1.8 The correlation coefficient between X and Y is closest to √ (a) – 0.96 (b) – 0.92 (c) – 0.88 (d) – 0.84 (e) – 0.80 6. Let A and B be two events, 6 . 0 ) | ( = B A P , P (at least one of the two events occurs) = 0.8, and P (exactly one of the two events occurs) = 0.6 where “exactly one” means either A or B , but not both A and B . Then P ( A ) is equal to (a) 2 1 (b) 3 1 (c) 4 3 √ (d) 3 2 (e) none of these

Page 6 of 9 Part II Show your work in each question. 18. (13 points) In a certain factory, machine A produces 35% of the CD’s, machine B produces 25% of the CD’s, and machine C produces 40%. Past results show that machine A is twice as likely to produce defective CD’s as machine B; machines C is three times as likely to produce defective CD’s as machine B. A randomly selected CD this factory is defective, what is the probability that it is produced by machine A? Solution: Method 1. Let A = event that a CD is produced by machine A B = event that a CD is produced by machine B C = event that a CD is produced by machine C D = a CD is defective c D = the complement of D . p = P (machine B produces defective CDs) = ( ) B D P p A D P 2 ) ( = p p A D P 7 . 0 2 ) 35 . 0 ( ) ( = =  P ( A )=0.35 p A D P c 2 1 ) | ( − = ) 2 1 ( 35 . 0 ) ( p A D P c − =  p B D P = ) | ( p B D P 25 . 0 ) ( =  25 . 0 ) ( = B P p B D P c − = 1 ) | ( ) 1 ( 25 . 0 ) ( p B D P c − =  p C D P 3 ) | ( = p p C D P 2 . 1 3 ) 4 . 0 ( ) ( = =  P ( C )=0.40 p C D P c 3 1 ) | ( − = ) 3 1 ( 4 . 0 ) ( p C D P c − =  ) ( ) | ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( C P C D P B P B D P A P A D P C D P B D P A D P D P + + =  +  +  = = 0.7 p + 0.25 p + 1.2 p = 2.15 p ( ) ( ) ( ) 325581395 . 0 43 14 215 70 15 . 2 7 . 0 | = = = =  = p p D P D A P D A P Method 2. Supplier D c D Total A 0.7 p 0.35 – 0.7 p 0.35 B 0.25 p 0.25 – 0.25 p 0.25 C 1.2 p 0.4 – 1.2 p 0.40 Total 2.15 p 1 – 2.15 p 1.00 Since p B D P = ) ( ,we have p B P B D P B D P 25 . 0 ) ( ) | ( ) ( = =  . Similarly we can calculate other joint probabilities in the above table. Hence ( ) ( ) ( ) 325581395 . 0 43 14 215 70 15 . 2 7 . 0 | = = = =  = p p D P D A P D A P

Page 9 of 9 21. (13 points) Suppose a coin is biased. The probability of showing a head is 1 3 , and the probability of showing a tail is 2 3 . This coin is rolled until a head shows. Assume that the outcomes of the rolls are independent. (a) (5 points) What is the probability that it will takes more than 2 rolls for that to happen? Solution: Let E be the event that it takes more than 2 rolls for the first head to appear. Method 1. P ( E ) = 1 – P (it takes 1 or 2 rolls for the first head to appear) =             −       − 3 1 3 2 3 1 1 = 9 4 Method 2. ( ) ( ) ,... , , TTTTH TTTH TTH P E P = = ( ) ( ) ( ) ... + + + TTTTH P TTTH P TTH P = ... 3 1 3 2 3 1 3 2 3 1 3 2 4 3 2 +             +             +             =         +       +       +       +             ... 3 2 3 2 3 2 1 3 1 3 2 3 2 2 =                   −             3 2 1 1 3 1 3 2 2 = ( ) 9 4 3 3 1 9 4 =             where we have used the geometric series r r r r − = + + + + 1 1 ... 1 3 2 , 1 0   r . (b) (8 points) Suppose it takes more than 2 rolls to get the first head, what is the probability that the first head appearing on an odd-numbered roll? Solution: Let E be the event that it takes more than 2 rolls for the first head to appear. From part (A), ( ) 9 4 = E P . Let O = the event that the first head appears on an odd-numbered roll. ( ) ( ) ( )   9 / 4 ,... , , | TTTTTTH TTTTH TTH P E P E O P E O P =  = =         +             +             +                   ... 3 1 3 2 3 1 3 2 3 1 3 2 4 9 6 4 2 =         +       +       +                   ... 3 2 3 2 1 3 1 3 2 4 9 4 2 2 =                     −                   2 2 3 2 1 1 3 1 3 2 4 9 = 5 3