Assignment 1 (1206A1w2008) Solutions

# Assignment 1 (1206A1w2008) Solutions - MGSC 1206.2...

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MGSC 1206.2 SOLUTIONS: Assignment #1 Winter 2008 TOTAL MARKS = 75 Due: 1 pm, Friday, January 18 th 7.3 #22. A single fair die is rolled. Find the probability of the event “getting a number greater than 2”. Let E = “getting a number greater than 2” S = {1,2,3,4,5,6} so n(S)=6. The numbers greater than 2 are 3, 4, 5, 6 so n(E) = 4. P(E) = n(E) / n(S) = 4/6 = 0.67 7.3 #24. A single fair die is rolled. Find the probability of the event “getting any number except 3”. Let E = “getting any number except 3” S = {1,2,3,4,5,6} so n(S)=6. The numbers other than 3 are 1, 2, 4, 5, 6 so n(E) = 5. P(E) = n(E) / n(S) = 5/6 = 0.83 7.3 #44. In 1998, funding for university research in the US totaled \$26.343 billion. Support came from various sources, as shown below. Source Amount (in billion \$) Federal government 15.558 State and local government 2.070 Industry 1.896 Academic institutions 4.979 Other 1.840 Find the probabilities that funds for a particular project came from each of the following sources: (a) federal government, (b) industry, (c) academic institutions. We will assume that the probability that a particular project had funds from a given source is proportional to the fraction of total funds coming from that source. a. P( funds came from federal government ) = 15.558 / 26.343 = 0.591 b. P( funds came from industry ) = 1.896 / 26.343 = 0.072 c. P( funds came from academic institutions ) = 4.979 / 26.343 = 0.189 Thus the probabilities of funds coming from federal government, industry, or academic institutions are 59.1%, 7.2% and 18.9% respectively. 1 2 pts 2 pts 3 pts

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7.4 #54. The table below shows the probabilities of a person accumulating specific amounts of credit card charges over a 12-month period. Find the probabilities that a person’s total charges during the period are the following: (a) \$500 or more (b) less than \$1000 (c) \$500 to \$2999 (d) \$3000 or more. (a) Let E = “total charges are \$500 or more”. P(E)= 1–P(E’) = 1 – [0.31+0.18] = 0.51 OR P(E) = 0.18 + 0.13 + 0.08 + 0.05 + 0.06 + 0.1 =0.51 The probability that total charges are \$500 or more is 51%. (b) Let E = “total charges are less than \$1000”. P(E) = 0.31 + 0.18 + 0.18 = 0.67 The probability that total charges are less than \$1000 is 67%. (c) Let E = “total charges are \$500 to \$2999”. P(E) = 0.18 + 0.13 + 0.08 = 0.39 The probability that total charges are \$500 to \$2999 is 39%. (d) Let E = “total charges are \$3000 or more”. P(E) = 0.05 + 0.06 + 0.01 = 0.12 The probability that total charges are \$3000 or more is 12%. 7.4 #64. The following data were gathered for 130 adult US workers: 55 were women, 3 women earned more than \$40,000; and 62 men earned \$40,000 or less. Find the probabilities that an individual is: (a) a woman earning \$40,000 or less (b) a man earning more than \$40,000 (c) a man or is earning more than \$40,000 (d) a woman or is earning \$40,000 or less. \$40000 or less
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Assignment 1 (1206A1w2008) Solutions - MGSC 1206.2...

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