Math 1030Q

Practice Final Exam

December 2008

Name and section: Instructor’s name: 1. 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:

Red & White Blue & White Blue & Gold Red & Gold

12 1X 2X 3 4

12 1X 3 2 4

Percentage of Voters 22 24 2X 2 1X 3 4 1X 3X 4

11 4 2X 1X 3

19 3 4 2X 1X

(a) Which choice would win using the plurality method? Solution: RW-24 BW-22 BG -35 RG -19 (b) Which choice would win in a runoff between the top two finishers of the plurality? Solution: RW-46 BG-54 (c) Which choice would win using Borda’s Method? Solution: RW-283 BW-248 BG -279 RG -190 (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-46 BW-45 BG -54 RG -41 (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. 2. 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:

Math 1030Q

Practice Final Exam, Page 2 of ?? Adama Laura Baltar

December 2008

20 27 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.

Math 1030Q

Practice Final Exam, Page 3 of ??

December 2008

3. A summer camp for children entering grades 2 through 6 employs 19 camp counselors. Each of the counselors will be in charge of a group of children all in the same grade. The number of children entering each grade is listed in the table. Grade # of Children

2 23

3 32

4 45

5 31

6 44

Apportion the camp counselors among the different grades using: (a) Hamilton’s method,

Grade 2nd 3rd 4th 5th 6th Total

Number of Children 23 32 45 31 44

Natural Quota D=

Initial Allocation

Final Allocation

(b) Lowndes’ method,

Grade 2nd 3rd 4th 5th 6th Total

Number of Children 23 32 45 31 44

Natural Quota D=

Initial Allocation

Relative Fractional Part

Final Allocation

(c) Jeffereson’s method,

Grade 2nd 3rd 4th 5th 6th Total

Number of Children 23 32 45 31 44 Grade 2nd 3rd 4th 5th 6th

Natural Quota D=

# of Children 23 32 45 31 44

Initial Allocation

# Seats

Modified Quota D=

Threshold Divisor

Final Allocation

Math 1030Q

Practice Final Exam, Page 4 of ??

December 2008

(d) Webster’s method,

Grade 2nd 3rd 4th 5th 6th Total

Number of Children 23 32 45 31 44 Grade 2nd 3rd 4th 5th 6th

Natural Quota D=

# of Children 23 32 45 31 44

Initial Allocation

# Seats

Modified Quota D=

Final Allocation

Threshold Divisor

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-3 3rd-3 4th-5 5th-3 6th-5 (B) 2nd-3 3rd-3 4th-5 5th-3 6th-5 (C) You can use the modified quota D = 8. 2nd-2 3rd-4 4th-5 5th-3 6th-5 (D) You can use the modified quota D = 9.2. 2nd-3 3rd-3 4th-5 5th-3 6th-5

Math 1030Q

Practice Final Exam, Page 5 of ??

4. Solve for x accurately to two decimal places: 2 − (9.2)−8x = 2.32 −14

Solution: x ≈ −.20. 5. Solve for y accurately to two decimal places:  3

y−7 6

13 = 1540

Solution: x ≈ 16.70.

December 2008

Math 1030Q

Practice Final Exam, Page 6 of ??

December 2008

6. Which will be worth more in 10 years: $6, 000 invested at 8.7% simple interest, or $6, 000 invested at 4.9% interest, compounded bi-monthly? Solution: For simple interest the future value is $11,200.00. For compound interest the future value is $9,774.43. 7. For an account with an annual interest rate of 6.8%, find the annual percentage yeild (APY) if interest is compounded: (a) quarterly (b) monthly (c) daily Solution: (A) APY = .06975, or 6.98%. (B) APY = .07015 or 7.01%. (B) APY = .07035 or 7.04%.

Math 1030Q

Practice Final Exam, Page 7 of ??

December 2008

8. 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. 9. 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.

Math 1030Q

Practice Final Exam, Page 8 of ??

December 2008

10. 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. 11. 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.

Math 1030Q

Practice Final Exam, Page 9 of ??

December 2008

12. Maggie borrows $7, 800 from the bank at 7.3% interest compounded monthly. If she makes a $400 payment at the end of the first month, how much does she then owe? Solution: She will owe $7,447.45 after her first payment. The method to solve this mimicks the similar question found on on test 2. You do not use the loan formula here. 13. Andrew takes out an $36, 100 student loan to pay for graduate school. If the interest rate is 4.7% compounded quarterly, how large would his quarterly payments be in order to pay off this loan in 8 years? Solution: Use the loan formula and solve for R. R = 1, 360.21.

Math 1030Q

Practice Final Exam, Page 10 of ??

December 2008

14. 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.

15. 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

P8

Solution: 15

(b)

100

P8 =

15! 15! = = 15 · 14 · 13 · · · 9 · 8 = 259459200 (15 − 8)! 7!

C70

Solution: 100

C70 =

100! 100 · 99 · 98 · · · 71 · 71 = = 2.937 × 1025 70!(100 − 70)! 30!

16. 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 ) =

51 37 + 23 − 9 = . 54 54

(b) have neither children nor pets? Solution: P (neither C nor P ) = 1 − P (C OR P ) =

3 . 54

Math 1030Q

Practice Final Exam, Page 11 of ??

December 2008

17. The Office of Admissions and Records of a large western university released the accompanying information concerning the contemplated majors of its freshman class:

Major Business Humanities Education Other

Percentage Freshmen Choosing This Major .29 .07 .18 .46

Percent of Females .38 .60 .67 .48

Percent of Males .52 .40 .33 .52

(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.

Math 1030Q

Practice Final Exam, Page 12 of ??

December 2008

18. Suppose you must sit 30 people at a round table (so which seat is first is not important) for Thanksgiving dinner. Of the 30 people, 10 are from your mom’s side of the family, 12 are from your dad’s side, and the remaining 8 are your neighbours. (a) In total, how many ways can you seat all 30 people at the table? Solution: Total ways to seat 30 people is 30! (b) What if you want to keep family members next to family members, so all your mom’s family is in one group, then your dad’s, then your neighbours? Solution: Keeping each family together we have (10!)(12!)(8!) ways. 19. From a bag containing six red, five white, and ten blue balls, three are drawn at random. What is the probability (a) that all three are the same colour? Solution: Since these are mutually exclusive events P (3R OR 3W OR 3B) = P (3R) + P (3W ) + P (3B) = .11278. 6 C3 . Exra hint: P (3R) = 21 C3 (b) that all three are different colours? Solution: P (all different) =

6 · 5 · 10 = .2255. 21 C3

(c) that at least two two of the three are the same colour? Solution: P (two same) = 1 − P (all different) = .7744.

Math 1030Q

Practice Final Exam, Page 13 of ??

December 2008

20. Consider a game that consists of drawing a single card at random from a standard deck of cards (52 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. 21. 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 0.00266 . (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.

Math 1030Q

Practice Final Exam, Page 14 of ??

December 2008

22. 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 o 2 . 4 1 (B) . 4 0 (C) . 4 1 (D) . 4 (A)

A AA Ao

B AB Bo

Math 1030Q

Practice Final Exam, Page 15 of ??

A Formula Sheet: Simple Interest: Compound Interest:

Systematic Savings:

Loans:

F = P (1 + rt)  r nt F =P 1+ n  r n AP Y = 1 + −1 n   r nt 1+ −1   n F = D  r n    r −nt 1− 1+   n P = R  r n

Bayes’ Theorem:

P (Ai |B) =

P (B|Ai )P (Ai ) P (B|A1 )P (A1 ) + · · · + P (B|An )P (An )

OR P (Ai |B) =

P (B|Ai )P (B) . P (B)

December 2008