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Math 1030Q Practice Final Exam December 2008
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Instructor’s name:
- A community votes to choose colours for its new high school. The choices are blue & white, red & white, blue & gold, and red & gold. Their preference rankings are as follows:
Percentage of Voters 12 12 22 24 11 19 Red & White 1 X 1 X 2 X 2 4 3 Blue & White 2 X 3 1 X 3 2 X 4 Blue & Gold 3 2 4 1 X 1 X 2 X Red & Gold 4 4 3 X 4 3 1 X
(a) Which choice would win using the plurality method? Solution: RW- BW- BG - RG -
(b) Which choice would win in a runoff between the top two finishers of the plurality? Solution: RW- BG-
(c) Which choice would win using Borda’s Method? Solution: RW- BW- BG - RG -
(d) Which choice if any would be the Condorcet’s winner? Solution: Blue and Gold wins all three of it’s head to head comparisions.
(e) Which choice would win in an approval vote? Solution: RW- BW- BG - RG -
(f) Can the people who voted for blue & gold first and red & white second (the column receiving 24% of the vote) obtain a preferable outcome in a Borda vote by changing their preference rankings?
Explain your reasoning clearly. If your answer is “yes”, indicate how the preferable outcome is obtained. If your answer is “no”, explain why no preferable outcome is possible. Solution: Yes. By switching their 4 and 2 choice.
- Suppose there are 115 votes cast in an election between three candidates - Adama, Laura, and Baltar - to be decided by a plurality. After the fist 85 votes are counted the tallies are as follows:
Adama 20 Laura 27 Baltar 38
(a) What is the minimal number of remaining votes Baltar needs to be assured of a win? Solution: Baltar needs 10 votes.
(b) What is the minimal number of remaining votes Laura needs to be assured of a win?
Solution: Laura needs 21 votes.
(d) Webster’s method,
Natural Modified Number of Quota Initial Quota Final Grade Children D= Allocation D= Allocation 2nd 23 3rd 32 4th 45 5th 31 6th 44 Total Grade # of Children # Seats Threshold Divisor 2nd 23 3rd 32 4th 45 5th 31 6th 44
Solution: The house size is 19 and the total number of children is 175, so the natrual quota is 9.210. Here are the final allocations: (A)
2nd- 3rd- 4th- 5th- 6th-
(B)
2nd- 3rd- 4th- 5th- 6th-
(C) You can use the modified quota D = 8.
2nd- 3rd- 4th- 5th- 6th-
(D) You can use the modified quota D = 9.2.
2nd- 3rd- 4th- 5th- 6th-
- Solve for x accurately to two decimal places:
2 − (9.2)−^8 x − 14
Solution: x ≈ −. 20.
- Solve for y accurately to two decimal places:
y − 7 6
Solution: x ≈ 16. 70.
- Ashley’s parents want to invest $10,000 now for her to go to college in 18 years. How much will money will they have if they deposit the money into an account earning 4.5% per year compounded monthly? (Your answer should be correct to two decimal places)
Solution: This is a basic compound interest problem. The future value of the account is $22,445.05.
- Suppose a friend lends you $103, and you agree to pay him back $124 in 19 months.
(a) If we assume that this is simple interest, then what is the interest rate? Solution: Use the simple interest formula and solve for r. r = .1287.
(b) If we assume that this is compounded monthly, then what is the interest rate? Solution: Use the compound interest formula and solve for r. r = .1178.
- I own $3, 500 of a stock which has grown at a rate of 6.8% annually. We want to estimate how long will it take for my investment to grow to $4, 200? To do this treat this as an account earning 6.8% compounded daily.
Solution: use the compound interest formula and solve for t. This will require the log rule. t = 2. 7 years or 32 months.
- Robin has determined that she would like to retire in 30 years. Right now she can make deposits of $460 per month into an account earning 3.8%, compounded monthly. How much money will Robin have when she retires?
Solution: Use the systematic savings formula and solve for F. F = 308, 123 .71.
- Two twelve sided dice are thrown. What is the probability that the sum of the two dice will not be 6?
Solution: There is a total of 144 possible rolls. Only 5 of which sum to six. Thus P ( sum to 6) = 5 /144. Hence, P ( not sum to 6) = 1 − P ( sum to 6) = 139/ 144.
- For each of the two parts, write down the expression involving factorials, that is the definition of the given permutation or combination number and then using cancellation and evaluation of factorials, find the corresponding numerical value. (a) 15 P 8 Solution: 15 P 8 =^
(b) 100 C 70 Solution: 100 C 70 =^
= 2. 937 × 1025
- In a particular neighborhood of 54 homes, 37 of the homes have children, 23 have pets, and 9 have both. What is the probability that a randomly selected home in the neighborhood will (a) have children or pets? Solution: P (C OR P ) =
(b) have neither children nor pets? Solution: P (neither C nor P ) = 1 − P (C OR P ) =
- The Office of Admissions and Records of a large western university released the accompanying informa- tion concerning the contemplated majors of its freshman class:
Percentage Freshmen Percent Percent Major Choosing This Major of Females of Males Business .29 .38. Humanities .07 .60. Education .18 .67. Other .46 .48.
(a) What is the probability that a student selected at random from the freshman class is female? (b) What is the probability that a business student selected at random from the freshman class is male? (c) What is the probability that a female student selected at random from the freshman class is majoring in business?
Solution: (A) P (F ) = .4936. (B) P (M ) = 1 − P (F ) = .5064. (C) Use Bayes’ Theorem, P (B|F ) = .2233.
- Consider a game that consists of drawing a single card at random from a standard deck of cards ( cards). You pay $3 to play the game, and the $ 3 is not returned. If you draw an ace you win $11. If you draw any other face card, you win $4. If you draw any other card, you win nothing. What is your expected value for the game?
Solution: Expected value is − 1 .2307.
- A man purchased a $250,000 one-year term life insurance policy for $750. On the basis of mortality rates for women of her age and background, the insurance company determines that the probability of the woman dying in the next year is 0.. (a) Find the insurance companys expected gain.
Solution: The expected gain is $85.
(b) How much should the insurance company charge to have an expected gain of $100? Solution: They should charge $765.
- There are four major blood types: A, B, AB, and O. Which blood type a person has is determined by a pair of genes, one comping from the mother and one from the father. The O gene is recessive, so the only way a person can have type O bloos is if they have the gene pair oo. The A and B types are both dominant over o. A person will have type A blood if they have either AA or Ao gene pairings. Similarly, the pairs BB and Bo result in type B blood. If a person has the gene pair AB, they he or she will have type AB blood. Mrs. Jung has type A blood, her father has type A, and her mother type O. If Mr. Jung has type AB blood, what is the probability that a child of Mr. and Mrs. Jung will have blood type (a) A? (b) B? (c) O? (d) AB?
Solution: You need the following table:
A B A AA AB o Ao Bo
(A)
(B)
(C)
(D)
