210-peters-07fafin-vB

# 210-peters-07fafin-vB - Math 210.— Topics in Finite Math...

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Unformatted text preview: Math 210 .— Topics in Finite Math Fall 2007 — Final Exam Name: TA + section nr: VERSION B a The exam has 6 problems. ( ) (b) Show your work for every problem. Answers without justiﬁcation will not receive credit. (c) Calculators are not allowed. ((1 There is a blank page and a normal random variable table at the end of the exam for if you need it. 1. (25 points) The average height of an inhabitant of 12th century France was 5 feet and about 30 percent of all inhabitants were taller than 5 feet and 3 inches. Estimate the standard deviation assuming that the height of the inhabitants is given by a normal random variable. [remember that 1 foot equals 12 inches] ’ 2. A mouse lives in a closed box with two rooms. If the mouse is currently in room 1 then the Chance that the mouse is still there after 1 minute is 1 / 4. If the mouse is currently in room 2 then the chance that the mouse is still there after 1 minute is 1/5. (a) (5 points) Give the transition matrix and transition diagram for this Markov chain. (b) (10 points) Suppose the mouse is currently in room 1. What is the chance that the mouse is in room 1 after 2 minutes? (c) (15 points) Estimate the chance that the mouse is in room '1 after 10 hours. 3. (25 points) A bank offers a special loan for start—up companies. They can loan 30,000 dollars and they won’t have to repay anything for the ﬁrst ﬁve years. Then they have to make sixty equal monthly payments for the next ﬁve years (the ﬁrst payment is made exactly 5 years after the loan is given). The sum of the present values of these monthly payments should equal the 30,000 dollars, the interest rate is 8 percent compounded monthly. ' How large should the monthly payments be‘.7 You are not allowed to use the formula for an annuity, but you are allowed to use the formula for a geometric sum. 4. (25 points) On January 1st of 2000 Jim opened a new savings account and since then deposited 100 dollars on the last day of every month. He also deposited an extra 1000 dollars on the last day of the years 2002 and 2005. If the interest rate is 6 percent compounded monthly, how much money was in his account on November 30th of 2007, immediately after Jim’s deposit? [If you use a formula you learned in class, please state it] 5. Eastern Paciﬁc College has a very interesting class schedule. EVery class meets once a week from 9am to 4pm (with a 1—hour break from noon to 1pm). For her ﬁrst semester Samantha will take one course each in mathematics, economy and biol— ogy. She can schedule the mathematics class on Monday, Tuesday or Wednesday, the economy class on Tuesday, Wednesday or Thursday and the biology class on Wednesday, Thursday or Friday. Of course, she can only take at most one class per day. (a) (15 points) How many possible schedules can she make? (b) (10 points) If she picks a schedule at random (each being equally likely), then what is the chance that she has a class on Wednesday? 6. Consider the Markov chain with initial state vector x0 = [1/2, 1/4, 1/4] and the following transi— tion matrix: H00 1 0 0 OHO (a) (10 points) Explain in English why this is not a regular Markov chain. (try to give a short but clear argument if possible) (b) (5 points) What is x2? (0) (10 points) What is \$2000? APPENDIX A AREAS UNDER THE STANDARD NORMAL CURVE A.1 USE OF THE TABLE The entries in Table A.1 are the areas under the standard normal curve from O to z. where the units and tenths digits of z are given in the left—hand column and the hundredths digit is given in the top line. For example, if z = 1.32, then ’ we look under 2 for the row labeled “1.3” and follow'that row to} the column 1.0 11 1.2 1.3 L4 1.5 1.6 1.7 1.8 1.9 20 21 22 23 2A 25 25 27 28 23 30 .3643 .3849 .4032 .4192 .4332 .4452 .4554 .4641 .4821 .4861 .4893 .4953 .4965 .4974 .4981 .3665 .3869 .4049 .4207- .4345 .4463 .4564 .4649 .4826 .4864 .4896 .4955 .4966 .4975 .4982 .3686 .3888 .4066 '.4222' .4357 .4474 .4573 .4656 .4830 .4868 .4898 .4956 .4967 .4976 .4982 .3708 .3907 .4082 .4236_ .4370 .4484 .4582 .4664 '.4834 .4871 .4901 .4957 .4968 .4977 :4983 .3729 .3925 ..4099 .4251 .4382 .4495 .4591 .4671 .4838 .4875 .4904 .4959 .4969 .4977 .4984 .3749 .3944 .4115 ,.4265 .4394 .4505 .4599 .4678 .4842 .4878 .4906 .4960 .4970 .4978 :4984 .3770 .3962 .4131 .4279 .4406 .4515 .4608 .4686 .4846 .4881 .4909 .4961 .4971 .4979 .4985 .3790 .3980 4147 . .4292 .4418 .4525 .4616 .4693 .4850 .4884 .4911 .4962 .4972 .4979 .4985 .4429 .4441 .4535 .4545 .4625 .4633 .4699 .4706 .4854 V .4857 .4887 .4890 .4913 .4916 .4963 .4973 .4980 .4986 labeled “.02.” The entry there tells us that the area under the standard normal curve from O to 1.32 is .4066. That is, Pr[0 5 Z 5 1.32] = .4066. TABLE A.1 Areas under the Standard Normal Curve z .00 .01 .02 .03 .04 .05 .06 .07 .08 ,09 00 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359 OJ .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753 02 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141 03 J179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 1517 Q4 4554 .1591 .1628 .1664 .1700 ' .1736 .1772; .1808 .1844 1879 05 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 2224 06 .2257 .2291 .2324 .2357 .2389 .2422 A.2454 .2486 .2517 .2549 07- .2580 .2611 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852 08 .2881 .2910. .2939 .2967 .2995 .3023 .3051 .3078 .3106 3133 09 .3810' _ .3830 .3997 .4015 .4162 .4177 14319 .4306 ‘ * ...
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