251y0421 3 24 04 ECO251 QBA1 SECOND HOUR EXAM March 24 2004 Name KEY Student Number Hour of Class Registered Circle 1pm 2pm Part I 48 points Do all the following All questions are 2 points each except as marked Exam is normed on 50 points including take home Bracketed numbers are a point total If you answer None of the above in any question you should provide an alternative answer and explain why You may receive credit for this even if you are wrong 1 If one event is not affected by the outcome of another event these two events can be regarded as a dependent events b independent events c mutually exclusive events d complementary events e none of the above Questions 2 5 are based on Exhibit 1 Show your work in questions 2 5 I should be able to tell what formula you are using Exhibit 1 A college bookstore has established that the distribution of book purchases for the fall term is as follows P x x 0 4 5 6 7 1 3 2 2 2 In other words there is a 10 probability that a student will buy no books etc Solution to 2 4 The table above is completed here Row P x F x xP x x x 2 P x 1 2 3 4 5 2 0 4 5 6 7 0 1 0 3 0 2 0 2 0 2 1 0 0 1 0 4 0 6 0 8 1 0 0 0 1 2 1 0 1 2 1 4 4 8 0 0 4 8 5 0 7 2 9 8 26 8 x E x2 2 P x 26 8 2 E x 2 2 26 8 4 8 2 3 76 1 40 1 1832 What is F 5 Cumulative distribution Answer 0 6 3 What is x E x Answer 4 8 4 What is Var x x2 Answer 3 76 5 xP x 4 8 E x If each book sells for 12 50 and y is the total expenditure on books Use only the results of questions 3 and 4 and 12 50 to find a y E y 1 Answer 60 b Var y 2 y 2 Answer 587 5 2 2 2 Explanation The formula says that if y ax y a x and y a x so if y 12 5 x a 12 5 Then y 12 5 x 12 5 4 8 60 and y2 a 2 x2 12 5 2 3 76 587 5 1 251y0421 3 24 04 6 Wathen A machine inserts mixed vegetables in a bag Results have been as below of Total Weight Underweight 2 5 Satisfactory 90 0 Overweight 7 5 We take a sample of 3 bags The probability that all are overweight is a 0000156 b 0004218 075 3 0004218 Actually 0004219 c 0000001 d 0750000 e 2250000 f none of the above 13 Questions 7 10 are based on exhibit 2 Exhibit 2 Edited from problems presented by Samuel Wathen for Lind et al 2002 The following joint probability table describes the Sales ability and the Potential for advancement of sales persons Advancement Potential Fair Good Excellent B3 B1 B2 Sales Ability Poor A1 Average A2 032 024 090 120 186 144 Good A3 044 090 270 If we complete the table by doing row and column sums we get 7 B2 B3 A1 024 044 100 A2 090 120 090 300 A3 Total 186 144 270 600 308 288 404 1 000 The probability that a salesperson selected at random has good sales ability and excellent potential for achievement is To 2 decimal places a 1 00 b 73 c 45 d 27 The table shows joint probabilities P A3 B3 270 e None of the above 15 8 Total B1 032 The probability that a salesperson selected at random has good sales ability or excellent potential for achievement is To 2 decimal places a 1 00 2 b P A3 B3 P A3 P B3 P A3 B3 600 404 270 73 734 c d e 45 27 None of the above 17 251y0421 3 24 04 9 The probability that a salesperson with good sales ability selected at random has excellent potential for achievement is To 2 decimal places a 1 00 b 73 P B3 A3 P B3 A3 270 45 Multiplication Rule P A3 6 c 45 d e 27 None of the above 19 10 Show your work P A1 B1 P A1 B1 032 1039 Multiplication Rule P B1 308 11 Fill in the following joint probability table on the assumption that a A1 and B1 are mutually exclusive and b A2 and B 2 are independent 4 25 B1 B2 A1 A2 A3 Total B3 20 20 40 Total 10 40 50 10 40 1 00 Solution First note that a if A1 and B1 are mutually exclusive then P A1 B1 0 and b if A2 and B 2 are independent P A2 B 2 P A2 P B 2 50 40 20 Fill these in A1 A2 A3 Total B1 B2 0 20 20 20 A1 following A2 A3 Total B3 Total 40 50 If you now fill in the rest to make it add up you get the 10 10 40 40 1 00 B1 B 2 B3 0 20 0 20 Total 20 20 0 20 10 10 40 50 10 40 40 1 00 12 A deck of cards consists of 52 cards There are 13 hearts in a deck You are dealt 6 cards a How many different possible hands are there b What is the chance of getting at least one heart You may leave your answer to b in factorial form 4 3 n Solution Remember C r 52 a So C 6 52 52 6 6 n n r r 52 46 6 52 51 50 49 48 47 2035620 6 5 4 3 2 1 4 251y0421 3 24 04 b The probability of getting no hearts is P 0 C 13 C 39 0 526 C6 13 39 13 0 33 6 52 46 6 So the probability of getting at least one heart is 1 P 0 1 C 013 C 639 C 652 13 39 13 0 33 6 1 39 46 1 Another way of doing this is 52 33 52 46 6 39 to recall that C 652 2035620 C 013 1 and that C 6 3262623 So P x 1 1 P 0 1 C 013 C 639 C 652 39 38 37 36 35 34 39 33 6 6 5 4 3 2 1 1 1 3262623 1 1603 8397 2035620 13 If you toss a coin 5 times in how many different ways can you get 2 heads 5 4 5 5 10 Solution C 2 3 2 2 1 14 If you toss a coin 5 times what is the probability that you get 2 heads Solution The probability of getting any given sequence of heads or tails is the same as that of getting 5 tails in a row …
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